4MeanValueTheorem 12-01-13

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52 CHAPTER 2. CALCULUS II

2.4 Mean Value Theorem

Course Objectives

• Roll’s Theorem

• Use Roll’s Theorem to prove that an equation has exactly one root

• Mean Value Theorem

• applications of Mean Value Theorem

Problems

1. Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Thenfind all numbers c that satisfy the conclusion of Rolle’s Theorem.

(a) f (x) = x2 − 4x + 1, [0, 4]

(b) f (x) = sin2πx, [−1, 1]

2. Let f (x) = 1 − x2

3 .

(a) Show that f (−1) = f (1).

(b) Show that there is no number c in (−1, 1) such that f (c) = 0.

(c) Why does this not contradict Rolle’s Theorem?

3. Let f (x) = (x − 1)−2.

(a) Show that f (0) = f (2).

(b) Show that there is no number c in (0, 2) such that f (c) = 0.

(c) Why does this not contradict Rolle’s Theorem?

4. Show that the equation x3 − 15x + 1 = 0 has at most one root in the interval [−2, 2].

5. Show that the equation x5 − 6x + 3 = 0 has at most one root in the interval [−1, 1].

6. Show that the equation x5 + 10x + 3 = 0 has exactly one real root.

7. Show that the equation 1 + 2x + x

3

+ 4x

5

= 0 has exactly one real root.8. Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval.

Then find all numbers c that satisfy the conclusion of the Mean Value Theorem.

(a) f (x) = 3x2 + 2x + 5, [−1, 1]

(b) f (x) = 3√

x, [0, 1]

9. Let f (x) = |x − 1|.

(a) Show that there is no number c such that f (c) = f (3) − f (0)

3 − 0 .

(b) Why does this not contradict the Mean Value Theorem?

10. Let f (x) = x + 1

x − 1 .

(a) Show that there is no number c such that f (c) = f (2) − f (0)

2 − 0 .

(b) Why does this not contradict the Mean Value Theorem?

11. If f (1) = 10 and f (x) ≥ 2 for 1 ≤ x ≤ 4, how small can f (4) possibly be?

12. Suppose that 3 ≤ f (x) ≤ 5 for all values of x. Show that 18 ≤ f (8) − f (2) ≤ 30.

13. Does there exist a function f such that f (0) = −1, f (2) = 4, and f (x) ≤ 2 for all x?



2.4. MEAN VALUE THEOREM 53

14. Prove the identity.

(a) tan−1 x + cot−1 x = π

2

(b) 2sin−1 x = cos−1(1 − 2x2) when x ≥ 0.

Bonus ProblemCorrectly done, you will receive 1 bonus point to the total final grade. Please turn it in on a separate pageof paper as the cover page of this homework assignment.

1. Use the Mean Value Theorem to prove the inequality: | sin a − sin b| ≤ |a − b| for all a and b.

Possible Quiz Questions

1. Show that the equation 3x − 2 + cos x = 0 has exactly one real root.

2. Suppose f is continuous on [2, 5], f (2) = −1 and 1 ≤ f (x) ≤ 4 for all x. Show that 2 ≤ f (5) ≤ 11.

3. Prove the identity tan−1 (x) + tan−1

1

x

=

π

2.

Answers

1. (a) i. f is continuous on [0, 4]ii. f is differentiable on (0, 4)

iii. f (0) = f (4) = 1

All the hypotheses are satisfied. So there should be a number c between 0 and 4 such thatf (c) = 0, which happens when c = 2.

(b) i. f is continuous on [−1, 1]

ii. f is differentiable on (−1, 1)

iii. f (−1) = f (1) = 0

All the hypotheses are satisfied. So there should be a number c between −1 and 1 such that

f (c) = 0, which happens when c = ±1

4 and c = ±3

4.

2. (a) f (−1) = f (1) = 0.

(b) f (x) = − 2

3 3√

x, which never equals 0.

(c) No, because f (0) doesn’t exist, i.e., f is not differentiable at all numbers in (−1, 1). One of thehypotheses fails to hold and Rolle’s Theorem does not apply.

3. (a) f (0) = f (2) = 1.

(b) f (x) = − 2

(x − 1)3, which never equals 0.

(c) No, because f (1) doesn’t exist, i.e., f is not continuous on [0, 2]. One of the hypotheses fails tohold and Rolle’s Theorem does not apply.

4. Let f (x) = x3

− 15x + 1. Assume the contrary, there exist two distinct roots a and b in (−2, 2), i.e.,f (a) = 0 and f (b) = 0. Now all the hypotheses of Rolle’s Theorem are satisfied.

(a) f is continuous on [a, b] (assume a > b without lost of generality).

(b) f is differentiable on (a, b).

(c) f (a) = f (b) = 0.

So there must be a number c between a and b such that f (c) = 0. But f (x) = 3x2 − 15, which isnever 0 when x is in (−2, 2). This is a contradiction. So our assumption that there are two distinctroots must be false, i.e., the equation x3 − 15x + 1 = 0 has at most one root in the interval [−2, 2].



54 CHAPTER 2. CALCULUS II

5. Let f (x) = x5 − 6x + 3. Assume the contrary, there exist two distinct roots a and b in (−1, 1), i.e.,f (a) = 0 and f (b) = 0. Now all the hypotheses of Rolle’s Theorem are satisfied.



(c) f (a) = f (b) = 0.

So there must be a number c between a and b such that f (c) = 0. But f (x) = 5x4− 6, which is never0 when x is in (

−1, 1). This is a contradiction. So our assumption that there are two distinct roots

must be false, i.e., the equation x5 − 6x + 3 = 0 has at most one root in the interval [−1, 1].

6. Define f (x) = x5 + 10x + 3. First, let’s show that f (x) = 0 has at least one root via the IntermediateValue Theorem.

(a) f is continuous.

(b) f (−1) = −8 and f (0) = 3

(c) 0 lies between −8 and 3

So by the IVT, there is at least one number c such that f (c) = 0, i.e., f (x) = 0 has at least one root.

Second, let’s show the uniqueness of the root. Assume the contrary, there exist two distinct roots a

and b, i.e., f (a) = 0 and f (b) = 0. Now all the hypotheses of Rolle’s Theorem are satisfied.



(c) f (a) = f (b) = 0.

So there must be a number c between a and b such that f (c) = 0. But f (x) = 5x4 + 10, which isnever 0. This is a contradiction. So our assumption that there are two distinct roots must be false.

Combining the two facts we have shown, we can conclude that there is exactly one solution.

7. Define f (x) = 1+2x+x3+4x5. First, let’s show that f (x) = 0 has at least one root via the IntermediateValue Theorem.

(a) f is continuous.

(b) f (−1) = −6 and f (0) = 1

(c) 0 lies between −6 and 1

So by the IVT, there is at least one number c such that f (c) = 0, i.e., f (x) = 0 has at least one root.

Second, let’s show the uniqueness of the root. Assume the contrary, there exist two distinct roots a

and b, i.e., f (a) = 0 and f (b) = 0. Now all the hypotheses of Rolle’s Theorem are satisfied.



(c) f (a) = f (b) = 0.

So there must be a number c between a and b such that f (c) = 0. But f (x) = 2 + 3x2 + 20x4, which

is never 0. This is a contradiction. So our assumption that there are two distinct roots must be false.

Combining the two facts we have shown, we can conclude that there is exactly one solution.

8. (a) i. f is continuous on [−1, 1]

ii. f is differentiable on (−1, 1)

Both hypotheses are satisfied. So there should be a number c between −1 and 1 such that

f (c) = f (1) − f (−1)

1 − (−1) = 2, which happens when c = 0.

(b) i. f is continuous on [0, 1]



2.4. MEAN VALUE THEOREM 55

ii. f is differentiable on (0, 1)

Both hypotheses are satisfied. So there should be a number c between 0 and 1 such that

f (c) = f (1) − f (0)

1 − 0 = 1, which happens when c =

√ 3

9 .

9. (a) f (3) − f (0)

3 − 0 =

1

3, but f (x) can’t be

1

3 (it is either 1 or −1).

(b) f is not differentiable at x = 0.

10. (a) f (2) − f (0)2 − 0 = 2, but f (x) = − 2(x − 1)2 can’t be 2 (it is always negative).

(b) f is not continuous at x = 1.

11. 16

12. Apply the Mean Value Theorem for f on [2, 8]. The condition 3 ≤ f (x) ≤ 5 for all values of x

immediate establishes that f is continuous and differentiable for all x. By the MVT, there exist c in

(2, 8) such that f (c) = f (8) − f (2)

8 − 2 . Thus 3 ≤ f (8) − f (2)

8 − 2 ≤ 5, i.e., 18 ≤ f (8) − f (2) ≤ 30.

13. No. According to the MVT, there would have to exist a number c such that f (c) = 5

2, which contradicts

the condition f (x) ≤ 2.

14. (a)

tan−1 x + cot−1 x

= 1

1 + x2 − 1

1 + x2 = 0. So tan−1 x + cot−1 x must be a constant. We can

decide what this constant is by plugging in one value of x. When x = 1, tan−1 1 + cot−1 1 = π

2.

Thus, tan−1 x + cot−1 x = π

2.

(b) when x ≥ 0,

2sin−1 x

= 2√ 1 − x2

,

cos−1(1 − 2x2)

= − 1

1 − (1 − 2x2)2· (−4x) =

4x√ 4x2 − 4x4

= 4x 4x2(1 − x2)

= 4x

2x√

1 − x2=

2√ 1 − x2

.

So (2sin−1 x) = (cos−1(1 − 2x2)) when x ≥ 0. This implies that 2 sin−1 x and cos−1(1 − 2x2)differ by a constant. We can decide what this constant is by plugging in x = 0. 2 sin−1 0 = 0,cos−1(1

−2

·02) = 0. Their difference is 0. Thus 2 sin−1 x = cos−1(1

−2x2) when x

≥0.

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4MeanValueTheorem 12-01-13