Chapter 23

Answers to Practice Problem Sets

CHAPTER 3 ANSWERS AND EXPLANATIONS

Practice Problem Set 1

  1. 13

To find the limit, we simply plug in 8 for x: (x² − 5 x − 11)= (8² − (5)(8) − 11) = 13.

  2.

To find the limit, we simply plug in 5 for x: = = = .

  3. π²

To find the limit, we would plug in π for x, but there is no x in the limit. So the limit is simply π².

  4. 4

If we plug in 3 for x, we get , which is indeterminate. When this happens, we try to factor the expression in order to get rid of the problem terms. Here we factor the top and get: = . Now we can cancel the term x − 3 to get (x+1). Notice that we are allowed to cancel the terms because x is not 3 but very close to 3. Now we can plug in 3 for x: (x+1) = 3 + 1 = 4.

  5. 0

Here we are finding the limit as x goes to infinity. We divide the top and bottom by the highest power of x in the expression: = . Next, simplify the top and bottom: . Now, if we take the limit as x goes to infinity, we get = = 0.

  6. + ∞

Here we are finding the limit as x goes to infinity. We divide the top and bottom by the highest power of x in the expression, which is x⁴: = . Next, simplify the top and bottom: . Now, if we take the limit as x goes to infinity, we get = = ∞.

  7.

Here we are finding the limit as x goes to infinity. We divide the top and bottom by the highest power of x in the expression, which is x⁴: = . Next, simplify the top and bottom: . Now, if we take the limit as x goes to infinity, we get = = .

  8.

Here we are finding the limit as x goes to infinity. We divide the top and bottom by the highest power of x in the expression, which is x². Notice that, under the radical, we divide by x⁴ because = x²: = . Next, simplify the top and bottom: . Now, if we take the limit as x goes to infinity, we get = = .

  9. +∞

Here we have to think about what happens when we plug in a value that is very close to 6, but a little bit more. The top expression will approach 8. The bottom expression will approach 0, but will be a little bit bigger. Thus, the limit will be , which is +∞.

10. −∞

Here we have to think about what happens when we plug in a value that is very close to 6, but a little bit less. The top expression will approach 8. The bottom expression will approach 0 but will be a little bit less. Thus, the limit will be , which is −∞.

11. The limit Does Not Exist.

In order to evaluate the limit as x approaches 6, we find the limit as it approaches 6⁺ (from the right) and the limit as it approaches 6^– (from the left). If the two limits approach the same value, or both approach positive infinity or both approach negative infinity, then the limit is that value, or the appropriately signed infinity. If the two limits do not agree, the limit “Does Not Exist.” Here, if we look at the solutions to problems 9 and 10, we find that as x approaches 6⁺, the limit is +∞, but as x approaches 6^–, the limit is −∞. Because the two limits are not the same, the limit Does Not Exist.

12. 1

Here we have to think about what happens when we plug in a value that is very close to 0, but a little bit more. The top and bottom expressions will both be positive and the same value, so we get .

13. −1

Here we have to think about what happens when we plug in a value that is very close to 0, but a little bit less. The top expression will be negative, and the bottom expression will be positive, so we get .

14. +∞

Here we have to think about what happens when we plug in a value that is very close to 7, but a little bit more. The top expression will approach 7. The bottom expression will approach 0 but will be a little bit positive. Thus, the limit will be , which is +∞.

15. The limit Does Not Exist.

In order to evaluate the limit as x approaches 7, we find the limit as it approaches 7⁺ (from the right) and the limit as it approaches 7^– (from the left). If the two limits approach the same value, or both approach positive infinity or both approach negative infinity, then the limit is that value, or the appropriately signed infinity. If the two limits do not agree, the limit “Does Not Exist.” Here, if we look at the solutions to problem 14, we see that as x approaches 7⁺, the limit is +∞. As x approaches 7^–, the top expression will approach 7. The bottom will approach 0, but will be a little bit negative. Thus, the limit will be , which is −∞. Because the two limits are not the same, the limit Does Not Exist.

16. +∞

In order to evaluate the limit as x approaches 7, we find the limit as it approaches 7⁺ (from the right) and the limit as it approaches 7^– (from the left). If the two limits approach the same value, or both approach positive infinity or both approach negative infinity, then the limit is that value, or the appropriately-signed infinity. If the two limits do not agree, the limit “Does Not Exist.” Here we see that as x approaches 7⁺, the top expression will approach 7. The bottom expression will approach 0 but will be a little bit positive. Thus, the limit will be , which is +∞. As x approaches 7^–, the top expression will again approach 7. The bottom will approach 0 but will be a little bit positive. Thus, the limit will be , which is +∞. Because the two limits are the same, the limit is +∞.

17. (a) 4; (b) 5; (c) The limit Does Not Exist.

(a) Notice that f(x) is a piecewise function, which means that we use the function f(x) = x² − 5 for all values of x less than or equal to 3. Thus, f(x) = 3² − 5 = 4.

(b) Here we use the function f(x) = x + 2 for all values of x greater than 3. Thus, = 3 + 2 = 5.

(c) In order to evaluate the limit as x approaches 3, we find the limit as it approaches 3⁺ (from the right) and the limit as it approaches 3^– (from the left). If the two limits approach the same value, or both approach positive infinity or both approach negative infinity, then the limit is that value, or the appropriately signed infinity. If the two limits do not agree, the limit “Does Not Exist.” Here, if we refer to the solutions in parts (a) and (b), we see that f(x) =4 and f(x) =5. Because the two limits are not the same, the limit Does Not Exist.

18. (a) 4; (b) 4; (c) 4

(a) Notice that is a piecewise function, which means that we use the function f(x) = x² − 5 for all values of x less than or equal to 3. Thus, = 3² − 5 = 4.

(b) Here we use the function f(x) = x + 1 for all values of x greater than 3. Thus, f(x) = 3 + 1 = 4.

(c) In order to evaluate the limit as x approaches 3, we find the limit as it approaches 3⁺ (from the right) and the limit as it approaches 3^– (from the left). If the two limits approach the same value, or both approach positive infinity or both approach negative infinity, then the limit is that value, or the appropriately signed infinity. If the two limits do not agree, the limit “Does Not Exist.” Here, if we refer to the solutions in parts (a) and (b), we see that f(x) = 4 and f(x) = 4. Because the two limits are the same, the limit is 4.

19.

Here, if we plug in for x, we get .

20. 0

Here, if we plug in 0 for x, we get .

21. 3

Remember Rule No. 1, which says that . If we want to find the limit of its reciprocal, we can write this as . Here, if we plug in 0 for x, we get = (3)(1) = 3.

22.

Remember Rule No. 4, which says that . Here . If we want to evaluate the limit the long way, first we divide the numerator and the denominator of the expression by x: . Next, we multiply the numerator and the denominator of the top expression by 3 and the numerator and the denominator of the bottom expression by 8. We get . Now, we can evaluate the limit: .

23.

Here, we can rewrite the expression as . Remember Rule No. 4, which says that . Here, . If we want to evaluate the limit the long way, we first divide the numerator and the denominator of the expression by x: . Next, we multiply the numerator and the denominator of the top expression by 7 and the numerator and the denominator of the bottom expression by 5. We get . Now, we can evaluate the limit: .

24. The limit Does Not Exist.

The value of sin x oscillates between −1 and 1. Thus, as x approaches infinity, sin x does not approach a specific value. Therefore, the limit Does Not Exist.

25. 0

Here, as x approaches infinity, approaches 0. Thus, .

26. 0

Here, use the trigonometric identity sin² x = 1 − cos² x to rewrite the bottom expression: . Next, we can break up the limit into . Remember that and that as well. Now we can evaluate the limit: = (1)(1)(0) = 0.

27.

We can break up the limit into . Remember Rule No. 4, which says that . Here . If we want to evaluate the limit the long way, we first divide the numerator and the denominator of the expressions by x: . Next, we multiply the numerator and the denominator of the top expression by 7 and the numerator and the denominator of the bottom expression by 11. We get . Now, we can evaluate the limit: .

28. 6

Notice that if we plug in 0 for h, we get , which is indeterminate. If we expand the expression in the numerator, we get . This simplifies to . Next, factor h out of the top expression: . Now, we can cancel the h and evaluate the limit to get (6+h) = 6 + 0 = 6.

29. cos x

Notice that if we plug in 0 for h, we get , which is indeterminate. Recall that the trigonometric formula sin (A + B) = sin A cos B + cos A sin B. Here, we can rewrite the top expression as . We can break up the limit into . Next, factor sin x out of the top of the left-hand expression: . Now, we can break this into separate limits: . The left-hand limit is . The right-hand limit is . Finally, combine the left-hand and right-hand limits: .

30.

Notice that if we plug in 0 for h, we get , which is indeterminate. If we combine the two expressions on top with a common denominator, we get . We can simplify the top expression, leaving us with: . Next, simplify the expression into: . We can cancel the h to get . Now, if we evaluate the limit we get .

CHAPTER 4 ANSWERS AND EXPLANATIONS

Practice Problem Set 1

  1. Yes. It satisfies all three conditions.

In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the following conditions:

Condition 1: f(c) exists.

Condition 2: f(x) exists.

Condition 3: f(x) = f(c)

Let’s test each condition.

f(2) = 9, which satisfies condition 1.

f(x) = 9 and f(x) = 9, so f(x) = 9, which satisfies condition 2.

f(x) = 9 = f(2), which satisfies condition 3. Therefore, f(x) is continuous at x = 2.

  2. No. It fails condition 3.

In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the following conditions:

Condition 1: f(c) exists.

Condition 2: f(x) exists.

Condition 3: f(x) = f(c)

Let’s test each condition.

f(3) = 29, which satisfies condition 1.

f(x) = 30 and f(x) = 30, so f(x) = 30, which satisfies condition 2.

But f(x) ≠ f(c). Therefore, f(x) is not continuous at x = 3 because it fails condition 3.

  3. No. It fails condition 1.

In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the following conditions:

Condition 1: f(c) exists.

Condition 2: f(x) exists.

Condition 3: f(x) = f(c)

Notice that the function is not defined at f(3). Therefore, f(x) is not continuous at x = 3 because it fails condition 1.

  4. No. It is discontinuous at any odd integral multiple of .

Recall that sec x = . This means that sec x is undefined at any value where cos x = 0, which are the odd multiples of . Therefore, sec x is not continuous everywhere.

  5. No. It is discontinuous at the endpoints of the interval.

Recall that sec x = . This means that sec x is undefined at any value where cos x = 0. Also recall that cos = 0 and = 0. Therefore, sec x is not continuous everywhere on the interval .

  6. Yes.

Recall that sec x = . This means that sec x is undefined at any value where cos x = 0. Also recall that cos = 0 and = 0. Therefore, sec x is continuous everywhere on the interval because the interval does not include the endpoints.

  7. The function is continuous for k = .

In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the following conditions:

Condition 1: f(c) exists.

Condition 2: f(x) exists.

Condition 3: f(x) = f(c)

We will need to find a value, or values, of k that enables f(x) to satisfy each condition.

Condition 1: f(4) = 0

Condition 2: f(x) = 0 and f(x) = 16k − 9. In order for the limit to exist, the two limits must be the same. If we solve 16k − 9 = 0, we get k = .

Condition 3: If we now let k = , f(x) = 0 = f(4). Therefore, the solution is k = .

  8. The function is continuous for k = 6 or k = − 1.

In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the following conditions:

Condition 1: f(c) exists.

Condition 2: f(x) exists.

Condition 3: f(x) = f(c)

We will need to find a value, or values, of k that enables f(x) to satisfy each condition.

Condition 1: f(−3) = k² − 5k

Condition 2: f(x) = 6 and f(x) = 6, so f(x) = 6.

Condition 3: Now we need to find a value, or values, of k such that f(x) = 6 = f(3). If we set k² − 5k = 6, we obtain the solutions k = 6 and k = −1.

  9. The removable discontinuity is at .

A removable discontinuity occurs when you have a rational expression with common factors in the numerator and denominator. Because these factors can be cancelled, the discontinuity is “removable.” In practical terms, this means that the discontinuity occurs where there is a “hole” in the graph. If we factor f(x) = , we get f(x) = . If we cancel the common factor, we get f(x) = . Now, if we plug in x = 3, we get f(x) = . Therefore, the removable discontinuity is at .

10. (a) 0; (b) 0; (c) 1; (d) 1; (e) Does Not Exist; (f) a jump discontinuity at x = −3; a removable discontinuity at x = −3 and an essential discontinuity at x = 5.

(a) If we look at the graph, we can see that f(x) = 0.

(b) If we look at the graph, we can see that f(x) = 0.

(c) If we look at the graph, we can see that f(x) = 1.

(d) If we look at the graph, we can see that f(x) = 1.

(e) f(3) Does Not Exist.

(f) There are three discontinuities: (1) a jump discontinuity at x = −3; (2) a removable discontinuity at x = 3; and (3) an essential discontinuity at x = 5.

CHAPTER 5 ANSWERS AND EXPLANATIONS

Practice Problem Set 1

  1. 5

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = 5x and x = 3. This means that f(3) = 5(3) = 15 and f(3 + h) = 5(3 + h) = 15 + 5h. If we now plug these into the definition of the derivative, we get f′(3) = . This simplifies to f′(3) = 5 = 5.

If you noticed that the function is simply the equation of a line, then you would have seen that the derivative is simply the slope of the line, which is 5 everywhere.

  2. 4

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = 4x and x = −8. This means that f(−8) = 4(−8) = −32 and f(−8 + h) = 4(−8 + h) = −32 + 4h. If we now plug these into the definition of the derivative, we get f′(−8) = . This simplifies to f′(−8) = = 4.

If you noticed that the function is simply the equation of a line, then you would have seen that the derivative is simply the slope of the line, which is 4 everywhere.

  3. 20

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here and f(x) = 2x² and x = 5. This means that f(5) = 2(5)² = 50 and f(5 + h) = 2(2 + h)² = 2(25+ 10h + h²) = 50 + 20h + 2h². If we now plug these into the definition of the derivative, we get f′(5) = . This simplifies to f′(5) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(5) = (20 + 2h). Now we take the limit to get f′(5) = (20 + 2h) = 20.

  4. −10

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here, f(x) = 5x² and x = −1. This means that f(−1) = 5(−1)² = 5 and f(−1 + h) = 5(−1 + h)² = 5(1 − 2h + h²) = 5 − 10h + 5h². If we now plug these into the definition of the derivative, we get f′(−1) = . This simplifies to f′(−1) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(−1) = (−10+5h). Now we take the limit to get f′(−1) = (−10+5h) = −10.

  5. 16x

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = 8x² and f(x + h) = 8(x + h)² = 8(x² + 2xh + h²) = 8x² + 16xh + 8h². If we now plug these into the definition of the derivative, we get f′(x). This simplifies to f′(x) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(x) = (16x+8h). Now we take the limit to get f′(x) = (16x+8h) = 16x.

  6. −20x

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = −10x² and f(x + h) = −10(x + h)² = −10(x² + 2xh + h²) = −10x² − 20xh − 10h². If we now plug these into the definition of the derivative, we get f′(x) = . This simplifies to f′(x) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(x) = (−20x−10h). Now we take the limit to get f′(x) = (−20x − 10h) = −20x.

  7. 40a

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = 20x² and x = a. This means that f(a) = 20a² and f(a + h) = 20(a + h)² = 20(a² + 2ah + h²) = 20a² + 40ah + 20h². If we now plug these into the definition of the derivative, we get f′(a) = . This simplifies to f′(a) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(a) = (40a+20h). Now we take the limit to get f′(a) = (40a + 20h) = 40a.

  8. 54

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here, f(x) = 2x³ and x = −3. This means that f(−3) = 2(−3)³ = −54 and f(−3 + h) = 2(−3 + h)³ = 2((−3)³ + 3(−3)²h + 3(−3)h² + h³) = −54 + 54h − 18h² + 2h³. If we now plug these into the definition of the derivative, we get f′(−3) = . This simplifies to f′(−3) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(−3) = (54 −18h+2h²). Now we take the limit to get f′(−3) = (54 −18h+2h²) = 54.

  9. −9x²

We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = −3x³. This means that f(x + h) = −3(x + h)³ = −3(x³ + 3x²h + 3xh² + h³) = −3x³ − 9x²h − 9xh² − 3h³. If we now plug these into the definition of the derivative, we get f′(x) = . This simplifies to f′(x) = . Now we can factor out the h from the numerator and cancel it with the h in the denominator: . Now we take the limit to get f′(x) = = −9x².

10. 4x³

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = x⁴. This means that f(x + h) = (x + h)⁴ = (x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴). If we now plug these into the definition of the derivative, we get . This simplifies to . Now we can factor out the h from the numerator and cancel it with the h in the denominator: . Now we take the limit to get = 4x³.

11. 5x⁴

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = x⁵.

This means that f(x + h) = (x + h)⁵ = (x⁵ + 5x⁴h + 10x³h² + 10x²h³ + 5xh⁴ + h⁵). If we now plug these into the definition of the derivative, we get . This simplifies to . Now we can factor out the h from the numerator and cancel it with the h in the denominator: . Now we take the limit to get = 5x⁴.

12.

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = 2 and x = 9. This means that f(9) = 2 = 6 and f(9 + h) = 2. If we now plug these into the definition of the derivative, we get . Notice that if we now take the limit, we get the indeterminate form . With polynomials, we merely simplify the expression to eliminate this problem. In any derivative of a square root, we first multiply the top and the bottom of the expression by the conjugate of the numerator and then we can simplify. Here the conjugate is . We get . This simplifies to . Now we can cancel the h in the numerator and the denominator to get . Now we take the limit: .

13.

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = 5 and x = 8. This means that f(8) = 5 = 20 and f(8 + h) = . If we now plug these into the definition of the derivative, we get . Notice that if we now take the limit, we get the indeterminate form . With polynomials, we merely simplify the expression to eliminate this problem. In any derivative of a square root, we first multiply the top and the bottom of the expression by the conjugate of the numerator and then we simplify. Here the conjugate is . We get . This simplifies to . Now we can cancel the h in the numerator and the denominator to get . Now, we take the limit: .

14.

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = sin x and x = . This means that and . If we now plug these into the definition of the derivative, we get . Notice that if we now take the limit, we get the indeterminate form . We cannot eliminate this problem merely by simplifying the expression the way that we did with a polynomial. Recall that the trigonometric formula sin(A + B) = sin A cos B + cos A sin B. Here, we can rewrite the top expression as . We can break up the limit into . Next, factor out of the top of the left-hand expression: . Now, we can break this into separate limits: . The left-hand limit is . The right-hand limit is . Therefore, the limit is .

15. –sin x

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = cos x and f(x + h) = cos (x + h). If we now plug these into the definition of the derivative, we get . Notice that if we now take the limit, we get the indeterminate form . We cannot eliminate this problem merely by simplifying the expression the way that we did with a polynomial. Recall that the trigonometric formula cos (A + B) = cos A cos B − sin A sin B. Here we can rewrite the top expression as . We can break up the limit into . Next, factor cos x out of the top of the left-hand expression: . Now, we can break this into separate limits: . The left-hand limit is . The right-hand limit is = sin x • 1 = sin x. Therefore, the limit is − sin x.

16. 2x + 1

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = x² + x and f(x + h) = (x + h)² + (x + h) = x² + 2xh + h² + x + h. If we now plug these into the definition of the derivative, we get . This simplifies to . Now we can factor out the h from the numerator and cancel it with the h in the denominator: . Now we take the limit to get f′(x) = = 2x + 1.

17. 3x² + 3

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = x³ + 3x + 2 and f(x + h) = (x + h)³ + 3(x + h) + 2 = x³ + 3x²h + 3xh² + h³ + 3x + 3h + 2.

If we now plug these into the definition of the derivative, we get f′(x) = = . This simplifies to . Now we can factor out the h from the numerator and cancel it with the h in the denominator: f′(x) = = . Now we take the limit to get f′(x) = = 3x² + 3.

18.

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = and f(x + h) = . If we now plug these into the definition of the derivative, we get . Notice that if we now take the limit, we get the indeterminate form . We cannot eliminate this problem merely by simplifying the expression the way that we did with a polynomial. Here we combine the two terms in the numerator of the expression to get . This simplifies to . Now we can cancel the factor h in the numerator and the denominator to get . Now we take the limit: .

19. 2ax + b

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here f(x) = ax² + bx + c and f(x + h) = a(x + h)² + b(x + h) + c = ax² + 2axh + ah² + bx + bh + c.

If we now plug these into the definition of the derivative, we get f′(x) = = . This simplifies to . Now we can factor out the h from the numerator and cancel it with the h in the denominator: . Now we take the limit to get = 2ax + b.

20.

We find the derivative of a function, f(x), using the definition of the derivative, which is . Here, f(x) = and that f(x + h) = . If we now plug these into the definition of the derivative, we get . Notice that if we now take the limit, we get the indeterminate form . We cannot eliminate this problem merely by simplifying the expression the way that we did with a polynomial. Here we combine the two terms in the numerator of the expression to get . This simplifies to f′(x) = = = . Now we can cancel the h in the numerator and the denominator to get . Now we take the limit: .

CHAPTER 6 ANSWERS AND EXPLANATIONS

Practice Problem Set 1

  1. 64x³ + 16x

First, expand (4x² + 1)² to get 16x⁴ + 8x² + 1. Now, use the Power Rule to take the derivative of each term. The derivative of 16x⁴ = 16(4x³) = 64x³. The derivative of 8x² = 8(2x) = 16x. The derivative of 1 = 0 (because the derivative of a constant is zero). Therefore, the derivative is 64x³ + 16x.

  2. 10x⁹ + 36x⁵ = 18x

First, expand (x⁵ + 3x)² to get x¹⁰ + 6x⁶ + 9x². Now, use the Power Rule to take the derivative of each term. The derivative of x¹⁰ = 10x⁹. The derivative of 6x⁶ = 6(6x⁵) = 36x⁵. The derivative of 9x² = 9(2x) = 18x. Therefore, the derivative is 10x⁹ + 36x⁵ + 18x.

  3. 77x⁶

Simply use the Power Rule. The derivative is 11x⁷ = 11(7x⁶) = 77x⁶.

  4. 80x⁹

Simply use the Power Rule. The derivative is 8x¹⁰ = 8(10x⁹) = 80x⁹.

  5. 54x² + 12

Use the Power Rule to take the derivative of each term. The derivative of 18x³ = 18(3x²) = 54x². The derivative of 12x = 12. The derivative of 11 = 0 (because the derivative of a constant is zero). Therefore, the derivative is 54x² + 12.

  6. 6x¹¹

Use the Power Rule to take the derivative of each term. The derivative of x¹² = 12x¹¹. The derivative of 17 = 0 (because the derivative of a constant is zero). Therefore, the derivative is: (12x¹¹) = 6x¹¹.

  7. −3x⁸ − 2x²

Use the Power Rule to take the derivative of each term. The derivative of x⁹ = 9x⁸. The derivative of 2x³ = 2(3x²) = 6x². The derivative of 9 = 0 (because the derivative of a constant is zero). Therefore, the derivative is −(9x⁸ + 6x²) = −3x⁸ − 2x².

  8. 0

Don’t be fooled by the power. π⁵ is a constant so the derivative is zero.

  9.

Use the Power Rule to take the derivative of each term. The derivative of . The derivative of . The derivative of (remember the shortcut that we showed you on this page). Therefore, the derivative is .

10.

Use the Power Rule to take the derivative of each term. The derivative of −8x⁻⁸ = −8(−8x⁻⁹) = 64x⁻⁹. The derivative of (remember the shortcut that we showed you on this page). Therefore, the derivative is .

11.

Use the Power Rule to take the derivative of each term. The derivative of 6x⁻⁷ = 6(−7x⁻⁸) = −42x⁻⁸. The derivative of (remember the shortcut that we showed you on this page). Therefore, the derivative is .

12.

Use the Power Rule to take the derivative of each term. The derivative of x⁻⁵ = −5x⁻⁶. To find the derivative of , we first rewrite it as x⁻⁸. The derivative of x⁻⁸ = −8x⁻⁹. Therefore, the derivative is −5x⁻⁶ − 8x⁻⁹ = .

13.

Use the Power Rule to take the derivative of each term. The derivative of (remember the shortcut that we showed you on this page). To find the derivative of , we first rewrite it as x⁻³. The derivative of x⁻³ = −3x⁻⁴. Therefore, the derivative is .

14. 216x² − 48x + 36

First, expand (6x² + 3)(12x − 4) to get 72x³ − 24x² + 36x − 12. Now, use the Power Rule to take the derivative of each term. The derivative of 72x³ = 72(3x²) = 216x². The derivative of 24x² = 24(2x) = 48x. The derivative of 36x = 36. The derivative of 12 = 0 (because the derivative of a constant is zero). Therefore, the derivative is 216x² − 48x + 36.

15. −6 − 36x² + 12x³ − 5x⁴ − 14x⁶

First, expand (3 − x − 2x³)(6 + x⁴) to get 18 − 6x − 12x³ + 3x⁴ − x⁵ − 2x⁷. Now, use the Power Rule to take the derivative of each term. The derivative of 18 = 0 (because the derivative of a constant is zero). The derivative of 6x = 6. The derivative of 12x³ = 12(3x²) = 36x². The derivative of 3x⁴ = 3(4x³) = 12x³. The derivative of x⁵ = 5x⁴. The derivative of 2x⁷ = 2(7x⁶) = 14x⁶. Therefore, the derivative is −6 − 36x² + 12x³ − 5x⁴ − 14x⁶.

16. 0

Don’t be fooled by the powers. Each term is a constant so the derivative is zero.

17.

First, expand to get . Next, rewrite the terms as: 4x⁻⁴ − 2x⁻⁵ − 6x⁻⁶. Now, use the Power Rule to take the derivative of each term. The derivative of 4x⁻⁴ = 4(−4x⁻⁵) = −16x⁻⁵. The derivative of 2x⁻⁵ = 2(−5x⁻⁶) = −10x⁻⁶. The derivative of 6x⁻⁶ = −36x⁻⁷. Therefore, the derivative is −16x⁻⁵ + 10x⁻⁶ = 36x⁻⁷ = .

18.

Use the Power Rule to take the derivative of each term. The derivative of (remember the shortcut that we showed you on this page). The derivative of = 0 (because the derivative of a constant is zero). Therefore, the derivative is .

19.

First, expand (x² + 8x − 4)(2x⁻² + x⁻⁴) to get 2 + 16x⁻¹ − 7x⁻² + 8x⁻³ − 4x⁻⁴. Now, use the Power Rule to take the derivative of each term. The derivative of 2 = 0 (because the derivative of a constant is zero). The derivative of 16x⁻¹ = 16(−1x⁻²) = −16x⁻². The derivative of 7x⁻² = 7(−2x⁻³) = −14x⁻³. The derivative of 8x⁻³ = 8(−3x⁻⁴) = −24x⁻⁴. The derivative of 4x⁻⁴ = 4(−4x⁻⁵) = −16x⁻⁵. Therefore, the derivative is −16x⁻² + 14x⁻³ − 24x⁻⁴ + 16x⁻⁵ = .

20. 0

The derivative of a constant is zero.

21. 3x² + 6x + 3

First, expand (x + 1)³ to get x³ + 3x² + 3x + 1. Now, use the Power Rule to take the derivative of each term. The derivative of x³ = 3x². The derivative of 3x² = 3(2x) = 6x. The derivative of 3x = 3. The derivative of 1 = 0 (because the derivative of a constant is zero). Therefore, the derivative is 3x² + 6x + 3.

22.

Use the Power Rule to take the derivative of each term. The derivative of (remember the shortcut that we showed you on this page). Rewrite as and as . The derivative of . The derivative of . Therefore, the derivative is .

23. 6x² + 6x − 14

First, expand x(2x + 7)(x − 2) to get x(2x² + 3x − 14) = 2x³ + 3x² − 14x. Now, use the Power Rule to take the derivative of each term. The derivative of 2x³ = 2(3x²) = 6x². The derivative of 3x² = 3(2x) = 6x. The derivative of 14x = 14. Therefore, the derivative is 6x² + 6x − 14.

24.

First, rewrite the terms as . Next, distribute to get: . Now, use the Power Rule to take the derivative of each term. The derivative of . The derivative of . Therefore, the derivative is .

25. 5ax⁴ + 4bx³ + 3cx² + 2dx + e

Use the Power Rule to take the derivative of each term. The derivative of ax⁵ = a(5x⁴) = 5ax⁴. The derivative of bx⁴ = b(4x³) = 4bx³. The derivative of cx³ = c(3x²) = 3cx². The derivative of dx² = d(2x) = 2dx. The derivative of ex = e. The derivative of f = 0 (because the derivative of a constant is zero). Therefore, the derivative is 5ax⁴ + 4bx³ + 3cx² + 2dx + e.

Practice Problem Set 2

  1.

We find the derivative using the Quotient Rule, which says that if , then . Here , so u = 4x³ − 3x² and v = 5x⁷ + 1. Using the Quotient Rule, we get . This can be simplified to .

  2. 3x² − 6x − 1

We find the derivative using the Product Rule, which says that if f(x) = uv, then . Here f(x) = (x² − 4x + 3)(x + 1), so u = x² − 4x + 3 and v = x + 1. Using the Product Rule, we get f′(x) = (x² − 4x + 3)(1) + (x + 1)(2x − 4). This can be simplified to f′(x) = 3x² − 6x − 1.

  3. 10(x + 1)⁹

We find the derivative using the Chain Rule, which says that if y = f(g(x)), then . Here f(x) = (x + 1)¹⁰. Using the Chain Rule, we get f′(x) = 10(x + 1)⁹(1) = 10(x + 1)⁹.

  4.

We find the derivative using the Chain Rule, which says that if y = f(g(x)), then . Here f(x) = , which can be written as f(x) = . Using the Chain Rule, we get . This can be simplified to .

  5.

Here, we will find the derivative using the Chain Rule. We will also need the Quotient Rule to take the derivative of the expression inside the parentheses. The Chain Rule says that if y = f(g(x)), then , and the Quotient Rule says that if f(x) = , then . We get . This can be simplified to .

  6.

Here, we will find the derivative using the Chain Rule. We will also need the Quotient Rule to take the derivative of the expression inside the parentheses. The Chain Rule says that if y = f(g(x)), then , and the Quotient Rule says that if f(x) = , then . We get . This can be simplified to .

  7.

We find the derivative using the Quotient Rule, which says that if f(x) = , then . Here , so u = 4x⁸ − and v = 8x⁴. Using the Quotient Rule, we get . This can be simplified to .

  8.

We have two ways that we could solve this. We could expand the expression first and then take the derivative of each term, or we could find the derivative using the Product Rule. Let’s do both methods just to see that they both give us the same answer. First, let’s use the Product Rule, which says that if f(x) = uv, then . Here , so and . Using the Product Rule, we get = . This can be simplified to + = .

The other way we could find the derivative is to expand the expression first and then take the derivative. We get . Now we can take the derivative of each term. We get . As we can see, the second method is a little quicker, and they both give the same result.

  9.

Here, we will find the derivative using the Chain Rule. We will also need the Quotient Rule to take the derivative of the expression inside the parentheses. The Chain Rule says that if y = f(g(x)), then , and the Quotient Rule says that if f(x) = , then . We get . This can be simplified to .

10. 100(x² + x)⁹⁹(2x + 1)

Here, we will find the derivative using the Chain Rule. The Chain Rule says that if y = f(g(x)), then . We get f′(x) = 100(x² + x)⁹⁹(2x + 1).

11.

Here, we will find the derivative using the Chain Rule. We will also need the Quotient Rule to take the derivative of the expression inside the parentheses. The Chain Rule says that if y = f(g(x)), then , and the Quotient Rule says that if f(x) = , then . We get . This can be simplified to .

12.

We find the derivative using the Quotient Rule, which says that if f(x) = , then . Here , so u = (x + 4)(x − 8) and v = (x + 6)(x − 6). Before we take the derivative, we can simplify the numerator and denominator of the expression: . Now using the Quotient Rule, we get . Next, we don’t simplify. We simply plug in x = 2 to get = .

13. 106

We find the derivative using the Quotient Rule, which says that if f(x) = , then . We will also need the Chain Rule to take the derivative of the expression in the denominator. The Chain Rule says that if y = f(g(x)), then , here , so u = x⁶ + 4x³ + 6 and v = (x⁴ − 2)². We get . Now, we don’t simplify. We simply plug in x = 1 to get .

14. 0

Here, we will find the derivative using the Chain Rule. We will also need the Quotient Rule to take the derivative of the expression inside the parentheses. The Chain Rule says that if y = f(g(x)), then , and the Quotient Rule says that if f(x) = , then . We get . Now we don’t simplify. We simply plug in x = 1 to get .

15.

We find the derivative using the Quotient Rule, which says that if f(x) = , then . Here, , so u = x² − 3 and v = x – 3. Using the Quotient Rule, we get . This can be simplified to .

16. 6

We find the derivative using the Product Rule, which says that if f(x) = uv, then . Here f(x) = (x⁴ − x²)(2x³ + x), so u = x⁴ − x² and v = 2x³ + x. Using the Product Rule, we get f′(x) = (x⁴ − x²)(6x² + 1) + (2x³ + x)(4x³ − 2x). Now we don’t simplify. We simply plug in x = 1 to get f′(x) = ((1)⁴ − (1)²)(6(1)² + 1) + (2(1)³ + (1))(4(1)³ − 2(1)) = 6.

17.

We find the derivative using the Quotient Rule, which says that if f(x) = , then . Here , so u = x² + 2x and v = x⁴ − x³. Using the Quotient Rule, we get . Now, we don’t simplify. We simply plug in x = 2 to get .

18.

We find the derivative using the Chain Rule which says that if y = f(g(x)), then . Here, f(x) = . Using the Chain Rule, we get . This can be simplified to .

19. −

We find the derivative using the Quotient Rule, which says that if f(x) = , then . We will also need the Chain Rule to take the derivative of the expression in the denominator. The Chain Rule says that if y = f(g(x)), then . Here , so u = x and v = (1 + x²)². We get . Now we don’t simplify. We simply plug in x = 1 to get .

20.

We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then . Here = 2u and = −1(x − 1)⁻² = . Thus, and because u = , .

21. −24

We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then . Here and = 3x². Now, we plug x = 1 into the derivative. Note that where x = 1, t = (1)³ = 1. We get and = 3(1)² = 3. Thus, .

22.

Here, the solution will be much simpler if we first substitute x = into the expression for y. We get: . Now, we find the derivative using the Product Rule, which says that if f(x) = uv, then . Here , so and v = 5t + . Using the Product Rule, we get .

23.

We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then . Here and . Now, we plug v = 2 into the derivative. Note that, where v = 2, x = . We get and , so .

24. 2

We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then . Here and = 2x. Now, we plug x = 1 into the derivative. Note that, where x = 1, u = 0. We get and = 2, so = (1)(2) = 2.

25.

We find the derivative using the Chain Rule, which says that if y = y(v) and v = v(x), then , although in this case we have u(y), y(x), and x(v), so we will find by . Here = 3y², , and = 2v. Next, . Now because x = v² and , we get .

Practice Problem Set 3

  1. 2 sin x cos x or sin 2x

Recall that . Here, we use the Chain Rule to find the derivative: . If you recall your trigonometric identities, 2 sin x cos x = sin 2x. Either answer is acceptable.

  2. −2x sin(x²)

Recall that . Here, we use the Chain Rule to find the derivative: .

  3. 2 sec³ x − sec x

Recall that and that . Using the Product Rule, we get . This can be simplified to sec³ x + sec x tan² x = 2 sec³ x − sec x.

  4. −4 csc²(4x)

Recall that . Here we use the Chain Rule to find the derivative: .

  5.

Recall that . Here we use the Chain Rule to find the derivative: . This can be simplified to .

  6.

Recall that . Here we use the Quotient Rule to find the derivative: . This can be simplified to .

  7. −4x csc²(x²) cot(x²)

Recall that . Here we use the Chain Rule to find the derivative: .

  8. 6 cox 3x cos 4x − 8 sin 3x sin 4x

Recall that and that . Here we use the Product Rule to find the derivative: . This can be simplified to .

  9. 16 sin 2x

Recall that and that . Here we will use the Chain Rule four times to find the fourth derivative.

The first derivative is: .

The second derivative is: .

The third derivative is: .

And the fourth derivative is: .

10. [cos(1 + cos² x) + sin(1 + cos² x)](−2 sin x cos x)

Recall that and that . Here we will use the Chain Rule to find the derivative: and . Next, because and t = 1 + cos² x, we get = (cos t + sin t)(-2 sin x cos x) = [cos(1 + cos² x) + sin(1 + cos² x)](-2 sin x cos x).

11.

Recall that . Using the Quotient Rule and the Chain Rule, we get . This simplifies to .

12. (sec θ)(sec² (2θ))(2) + (tan 2θ)(sec θ tan θ)

Recall that and that . Using the Product Rule and the Chain Rule, we get .

13. −[sin(1 + sin θ)](cos θ)

Recall that and that . Using the Chain Rule, we get .

14.

Recall that and that . Using the Quotient Rule, we get . This can be simplified (using trigonometric identities) to .

15.

Recall that . Here we use the Chain Rule to find the derivative: .

16.

Recall that and that . Using the Chain Rule, we get .

CHAPTER 7 ANSWERS AND EXPLANATIONS

Practice Problem Set 1

  1.

We take the derivative of each term with respect to x: .

Next, because = 1, we can eliminate that term and get: .

Next, group the terms containing .

Factor out the term . Now, we can isolate .

  2.

We take the derivative of each term with respect to x: .

Next, because = 1 we can eliminate that term and we can distribute the −16 to get .

Next, group the terms containing on one side of the equal sign and the other terms on the other side: .

Factor out the term . Now we can isolate , which can be reduced to .

  3.

First, cross-multiply so that we don’t have to use the Quotient Rule: x + y = 3x − 3y. Next, simplify 4y = 2x, which reduces to y = x. Now, we can take the derivative: . Note that just because a problem has the x’s and y’s mixed together doesn’t mean that we need to use implicit differentiation to solve it!

  4.

We take the derivative of each term with respect to x: .

Next, because = 1 we can eliminate that term to get .

Next, group the terms containing on one side of the equal sign and the other terms on the other side: .

Factor out the term . Now, we can isolate : , which can be simplified to .

  5.

We take the derivative of each term with respect to x: .

Next, because = 1 we can eliminate that term to get . Next, don’t simplify. Plug in (1, 1) for x and y: , which simplifies to .

Finally, we can solve for .

  6.

We take the derivative of each term with respect to x:

Next, because = 1 we can eliminate that term to get . Next, don’t simplify. Plug in (1, 1) for x and y, , which simplifies to .

Finally, we can solve for .

  7. −1

We take the derivative of each term with respect to x: .

Next, because = 1 we can eliminate that term to get . Next, don’t simplify. Plug in for x and y: , which simplifies to . If we multiply through by we get .

Finally, we can solve for .

  8.

We take the derivative of each term with respect to x: .

Next, because = 1 we can eliminate that term to get . Next, we can isolate . Now, we take the derivative again: . Next, because = 1 and we get . This can be simplified to .

  9.

We take the derivative of each term with respect to x: .

Next, because = 1 we can eliminate that term to get . Next, we can isolate . Now, we take the derivative again: . Next, because = 1 and we get . This can be simplified to .

10. 1

We can easily isolate y in this equation: y = x² − 2x + 1. We take the derivative: = x − 2. And we take the derivative again: . Note that just because a problem has the x’s and y’s mixed together doesn’t mean that we need to use implicit differentiation to solve it!

CHAPTER 8 ANSWERS AND EXPLANATIONS

Practice Problem Set 1

  1. y − 2 = 5(x − 1)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 1 into the equation y = 3x² − x, and we find the slope by plugging x = 1 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = 3(1)² − 1 = 2. This means that the line passes through the point (1, 2).

Next, we take the derivative: . Now, we can find the slope, m: . Finally, we plug in the point (1, 2) and the slope m = 5 to get the equation of the tangent line: y − 2 = 5(x − 1).

  2. y − 18 = 24(x − 3)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 3 into the equation y = x³ − 3x, and we find the slope by plugging x = 3 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = (3)³ − 3(3) = 18. This means that the line passes through the point (3, 18).

Next, we take the derivative: = 3x² − 3. Now, we can find the slope, m: = 3(3)² − 3 = 24. Finally, we plug in the point (3, 18) and the slope m = 24 to get the equation of the tangent line: y − 18 = 24(x − 3).

  3. y − 4 = −(x − 2)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 2 into the equation y = , and we find the slope by plugging x = 2 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = = 4. This means that the line passes through the point (2, 4).

Next, we take the derivative: = . Now, we can find the slope, m: = 1. However, this is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the tangent line’s slope. In this case, the slope of the normal line is = −1. Finally, we plug in the point (2, 4) and the slope m = −1 to get the equation of the normal line: y − 4 = −(x − 2).

  4. y −(x − 3)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 3 into the equation y = , and we find the slope by plugging x = 3 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = . This means that the line passes through the point (3, ).

Next, we take the derivative: . Now, we can find the slope, m: . Finally, we plug in the point (3, ) and the slope m = to get the equation of the tangent line: y − = (x − 3).

  5. y − 7 = (x − 4)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 4 into the equation y = , and we find the slope by plugging x = 4 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = = 7. This means that the line passes through the point (4, 7).

Next, we take the derivative: = . Now, we can find the slope, m: = −6. However, this is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the tangent line’s slope. In this case, the slope of the normal line is . Finally, we plug in the point (4, 7) and the slope m = to get the equation of the normal line: y − 7 = (x − 4).

  6. y = −3x + 4

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the slope by plugging x = 0 into the derivative of the equation y = 4 − 3x − x². First, we take the derivative: = −3 − 2x. Now, we can find the slope, m: = −3−3(0)=−3. Finally, we plug in the point (0, 4) and the slope m = −3 to get the equation of the tangent line: y − 4 = −3(x − 0) or y = −3x + 4.

  7. y = 0

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 2 into the equation y = 2x³ − 3x² − 12x + 20, and we find the slope by plugging x = 2 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = 2(2)³ − 3(2)² − 12(2) + 20 = 0. This means that the line passes through the point (2, 0).

Next, we take the derivative: = 6x² − 6x − 12. Now, we can find the slope, m: = 6 (2)² − 6(2)-12 = 0. Finally, we plug in the point (2, 0) and the slope m = 0 to get the equation of the tangent line: y − 0 = 0(x − 2) or y = 0.

  8. y + 29 = −39(x − 5)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 5 into the equation y = , and we find the slope by plugging x = 5 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = = −29. This means that the line passes through the point (5, −29).

Next, we take the derivative: . Now, we can find the slope, m: = = −39. Finally, we plug in the point (5, −29) and the slope m = −39 to get the equation of the tangent line: y + 29 = −39(x − 5).

  9. y − 7 = (x − 4)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the slope by plugging x = 4 into the derivative of the equation y = . First, we take the derivative: . Now, we can find the slope, m: . Finally, we plug in the point (4, 7) and the slope m = to get the equation of the tangent line: y − 7 = (x − 4).

10. y = 0

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = −2 into the equation y = (x² + 4x + 4)², and we find the slope by plugging x = −2 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = ((−2)² + 4(−2) + 4)². This means that the line passes through the point (−2, 0).

Next, we take the derivative: = 2(x² + 4x + 4)(2x + 4). Now, we can find the slope, m: = ((−2)² + 4(−2)+4)(2(−2)+4)=0. Finally, we plug in the point (−2, 0) and the slope m = 0 to get the equation of the tangent line: y − 0 = 0(x + 2) or y = 0.

11. x =

The slope of the line y = x is 1, so we want to know where the slope of the tangent line is equal to 1. We find the slope of the tangent line by taking the derivative: = 6x² − 8. Now we set the derivative equal to 1: 6x² − 8 = 1. If we solve for x, we get x = .

12. y − 7 = (x − 3)

Remember that the equation of a line through a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁). We find the y-coordinate by plugging x = 3 into the equation y = , and we find the slope by plugging x = 3 into the derivative of the equation.

First, we find the y-coordinate, y₁: y = = 7. This means that the line passes through the point (3, 7).

Next, we take the derivative: = . Now, we can find the slope, m: = −2. However, this is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the tangent line’s slope. In this case, the slope of the normal line is . Finally, we plug in the point (3, 7) and the slope m = to get the equation of the normal line: y − 7 = (x − 3).

13. x = 9

A line that is parallel to the y-axis has an infinite (or undefined) slope. In order to find where the normal line has an infinite slope, we first take the derivative to find the slope of the tangent line: = 2(x − 9)(1) = 2x − 18. Next, because the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line: m = . Now, we need to find where the slope is infinite. This is simply where the denominator of the slope is zero: x = 9.

14.

A line that is parallel to the x-axis has a zero slope. In order to find where the tangent line has a zero slope, we first take the derivative: = −3 − 2x. Now we need to find where the slope is zero. The derivative −3 − 2x = 0 at x = −. Now we need to find the y-coordinate, which we get by plugging x = − into the equation for y: 8 − 3. Therefore, the answer is .

15. a = 1, b = 0, and c = 1.

The two equations will have a common tangent line where they have the same slope, which we find by taking the derivative of each equation. The derivative of the first equation is: = 2x + a. The derivative of the second equation is = c + 2x. Setting the two derivatives equal to each other, we get: a = c. Each equation will pass through the point (−1, 0). If we plug (−1, 0) into the first equation, we get 0 = (−1)² + a(−1) + b, which simplifies to: a − b = 1. If we plug (−1, 0) into the second equation, we get 0 = c(−1) + (−1)², which simplifies to c = 1. Now we can find the values for a, b, and c. We get a = 1, b = 0, and c = 1.

Practice Problem Set 2

  1. c = 0

The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), then there exists at least one number c on the interval (a, b) such that . Here the function is f(x) = 3x² + 5x – 2 and the interval is [−1, 1]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′(c) = 5. Next, we need to find f′(c). The derivative of f(x) is f′(x) = 6x + 5, so f′(c) = 6c + 5. Now, we can solve for c: 6c + 5 = 5 and c = 0. Note that 0 is in the interval (−1, 1), just as we expected.

  2. c=

The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), then there exists at least one number c on the interval (a, b) such that f′(c) = . Here, the function is f(x) = x³ + 24x − 16 and the interval is [0, 4]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′(c) = 40. Next, we need to find f′(c) from the equation. The derivative of f(x) is f′(x) = 3x² + 24, so f′(c) = 3c² + 24. Now, we can solve for c: 3c² + 24 = 40 and c = . Note that is in the interval (0, 4), but − is not in the interval. Thus, the answer is only c = . It’s very important to check that the answers you get for c fall in the given interval when doing Mean Value Theorem problems.

  3. c = ≈ 0.62

The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), then there exists at least one number c on the interval (a, b) such that f′(c)=. Here, the function is f(x) = x³ + 12x² + 7x and the interval is [−4, 4]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′(c) = 23. Next, we need to find f′(c). The derivative of f(x) is f′(x) = 3x² + 24x + 7, so f′(c) = 3c² + 24c + 7. Now, we can solve for c: 3c² + 24c + 7 = 23 and c = . Note that c = is in the interval (−4, 4), but is not in the interval. Thus, the answer is only c = ≈ 0.62. It’s very important to check that the answers you get for c fall in the given interval when doing Mean Value Theorem problems.

  4. c =

The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), then there exists at least one number c on the interval (a, b) such that . Here the function is f (x) = − 3 and the interval is [1, 2]. Thus, the Mean Value Theorem says that f′(c) = . This simplifies to f′(c) = −3. Next, we need to find f′(c) from the equation. The derivative of f(x) is f′(x) = , so . Now, we can solve for c: = −3 and c = ± . Note that c = is in the interval (1, 2), but − is not in the interval. Thus, the answer is only c = . It’s very important to check that the answers you get for c fall in the given interval when doing Mean Value Theorem problems.

  5. No Solution.

The Mean Value Theorem says that: If f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), then there exists at least one number c on the interval (a, b) such that f′(c) = . Here, the function is f(x) = −3 and the interval is [−1, 2]. Note that the function is not continuous on the interval. It has an essential discontinuity (vertical asymptote) at x = 0. Thus, the Mean Value Theorem does not apply on the interval, and there is no solution.

Suppose that we were to apply the theorem anyway. We would get f′(c) = . This simplifies to f′(c) = 3. Next, we need to find f′(c) from the equation. The derivative of f(x) is f′(x) = , so . Now, we can solve for c: − = 3. This has no real solution. Therefore, remember that it’s very important to check that the function is continuous and differentiable everywhere on the given interval (it does not have to be differentiable at the endpoints) when doing Mean Value Theorem problems. If it is not, then the theorem does not apply.

  6. c = 4

Rolle’s theorem says that if f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), and if f(a) = f(b) = 0, then there exists at least one number c on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = x² – 8x + 12 and the interval is [2, 6]. First, we check if the function is equal to zero at both of the endpoints: f(6) = (6)² – 8(6) + 12 = 0 and f(2) = (2)² – 8(2) + 12 = 0. Next, we take the derivative to find f′(c): f′(x) = 2x – 8, so f′(c) = 2c – 8. Now, we can solve for c: 2c – 8 = 0 and c = 4. Note that 4 is in the interval (2, 6), just as we expected.

  7. c =

Rolle’s theorem says that if f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), and if f(a) = f(b) = 0, then there exists at least one number c on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = x³ – x and the interval is [−1, 1]. First, we check if the function is equal to zero at both of the endpoints: f(−1) = (−1)³ – (−1) = 0 and f(1) = (1)³ – (1) = 0. Next, we take the derivative to find f′(c): f′(x) = 3x² – 1, so f′(c) = 3c² – 1. Now, we can solve for c: 3c² – 1 = 0 and c = . Note that are both in the interval (−1, 1), just as we expected.

  8. c =

Rolle’s theorem says that if f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), and if f(a) = f(b) = 0, then there exists at least one number c on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = x(1 – x) and the interval is [0, 1]. First, we check if the function is equal to zero at both of the endpoints: f(0) = (0)(1 – 0) = 0 and f(1) = (1)(1 – 1) = 0. Next, we take the derivative to find f′(c): f′(x) = 1 – 2x, so f′(c) = 1 – 2c. Now, we can solve for c: 1 – 2c = 0 and c = . Note that is in the interval (0, 1), just as we expected.

  9. No Solution.

Rolle’s theorem says that if f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), and if f(a) = f(b) = 0, then there exists at least one number c on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = 1 − , and the interval is [−1, 1]. Note that the function is not continuous on the interval. It has an essential discontinuity (vertical asymptote) at x = 0. Thus, Rolle’s theorem does not apply on the interval, and there is no solution.

Suppose we were to apply the theorem anyway. First, we check if the function is equal to zero at both of the endpoints: f (1) = 1 − = 0 and f (−1) = 1 − = 0. Next, we take the derivative to find f′(c): , so f′(c) = . This has no solution.

Therefore, remember that it’s very important to check that the function is continuous and differentiable everywhere on the given interval (it does not have to be differentiable at the endpoints) when doing Rolle’s theorem problems. If it is not, then the theorem does not apply.

10. c =

Rolle’s theorem says that if f(x) is continuous on the interval [a, b] and is differentiable everywhere on the interval (a, b), and if f(a) = f(b) = 0, then there exists at least one number c on the interval (a, b) such that f(c) = 0. Here, the function is f(x) = and the interval is [0, 1]. First, we check if the function is equal to zero at both of the endpoints: f(0) = = 0 and f(1) = = 0. Next, we take the derivative to find f′(c) = , so f(c) = . Now, we can solve for c: = 0 and c = . Note that is in the interval (0, 1), just as we expected.