Mean Value Theorem

On this placeholder page we present the essentials of using the Mean Value Theorem and Rolle's Theorem. Then below we have typical problems, each with a complete solution immediately available, so you can see how they are used routinely, including in some typical exam questions. Let's jump right in!

MATHENO ESSENTIALS: Mean Value Theorem & Rolle's Theorem

Mean Value Theorem

If $𝑓 (𝑥)$ is continuous on the closed interval $[𝑎, 𝑏]$ and differentiable on the open interval $(𝑎, 𝑏)$ , then there is a number $𝑐$ in $(𝑎, 𝑏)$ such that
$𝑓' (𝑐) = 𝑓 (𝑏) - 𝑓 (𝑎) 𝑏 - 𝑎$
or, equivalently,
$𝑓 (𝑏) - 𝑓 (𝑎) = 𝑓' (𝑐) (𝑏 - 𝑎)$

In words: there is at least one value $𝑐$ between $𝑎$ and $𝑏$ where the tangent line is parallel to the secant line that connects the interval's endpoints. See the figures.

When the Mean Value Theorem applies, the slope of the tangent line at x=c is the same as the slope of the secant line connecting the endpoints of the interval.

Rolle's Theorem

In Calculus texts and lecture, Rolle's theorem is given first since it's used as part of the proof for the Mean Value Theorem (MVT). You can easily remember it, though, as just a special case of the MVT: it has the same requirements about continuity on $[𝑎, 𝑏]$ and differentiability on $(𝑎, 𝑏)$ , and the additional requirement that $𝑓 (𝑎) = 𝑓 (𝑏)$ . In that case, the MVT says that
$𝑓 (𝑏) - 𝑓 (𝑎) = 𝑓' (𝑐) (𝑏 - 𝑎) 0 = 𝑓' (𝑐) (𝑏 - 𝑎) 𝑓' (𝑐) = 0 f o r s o m e n u m b e r 𝑐 i n t h e o p e n i n t e r v a l (𝑎, 𝑏)$ See the figure.

The problems below illustrate some typical uses of the Mean Value Theorem and Rolle's Theorem.

Practice Problem #1: Straightforward Application of Rolle's Theorem

Consider the function

𝑓 (𝑥) = 9 - (𝑥 - 3) 2

on the interval

[0, 6]

.
(Note that

𝑓 (0) = 𝑓 (6) = 0

.)
Find the value(s) of

𝑐

that satisfy Rolle's Theorem.

Practice Problem #2: Straightforward Application of MVT

Consider the function

𝑓 (𝑥) = 𝑥 2

on the interval

[1, 4]

.
Find the value(s) of

𝑐

that satisfy the Mean-Value Theorem.

Practice Problem #3: How large can

𝑓 (𝑏)

be? (Based on a university exam question)

Suppose that

𝑓 (1) = 2

, and that

𝑓' (𝑥) \leq 3

for all values of

𝑥

. How large can

𝑓 (5)

possibly be?
[ Hint: Use the Mean Value Theorem.]

Practice Problem #4: Prove if

𝑓' (𝑥) > 0

, then

𝑓 (𝑥)

is an increasing function

Use the Mean Value Theorem to prove that if

𝑓 (𝑥)

is differentiable and

𝑓' (𝑥) > 0

for all

𝑥

, then

𝑓 (𝑥)

is an increasing function.

Practice Problem #5: Prove the following statements

Prove the following statements:

(a)

Use the Mean Value Theorem to prove that if

𝑓' (𝑥) = 0

for all

𝑥

in an interval

(𝑎, 𝑏)

, then

𝑓

is constant on

(𝑎, 𝑏)

(b)

Use the Mean Value Theorem to prove that if

𝑓' (𝑥) = 𝑔' (𝑥)

for all

𝑥

in an interval

(𝑎, 𝑏)

, then

𝑓' (𝑥) = 𝑔' (𝑐) + 𝐶

(𝑎, 𝑏)

, where

𝐶

is some constant.

[Hint: Consider the function

ℎ (𝑥) = 𝑓 (𝑥) - 𝑔 (𝑥)

, and use the result of part (a).]

Practice Problem #6: Prove that

| s i n (𝑎) - s i n (𝑏) | \leq | 𝑏 - 𝑎 |

Use the Mean Value Theorem to prove that

| s i n (𝑎) - s i n (𝑏) | \leq | 𝑏 - 𝑎 |

for all real numbers

𝑎

and

𝑏

Practice Problem #7: Prove that

| c o s (2 𝑎) - c o s (2 𝑏) | \leq 2 | 𝑏 - 𝑎 |

Use the Mean Value Theorem to prove that

| c o s (2 𝑎) - c o s (2 𝑏) | \leq 2 | 𝑏 - 𝑎 |

for all real numbers

𝑎

and

𝑏

Practice Problem #8: Prove that

𝑒 𝑥 > 1 + 𝑥

for

𝑥 > 0

(based on an actual university exam question)

Use the Mean Value Theorem to prove that

𝑒 𝑥 > 1 + 𝑥

for all

𝑥 > 0

The problem says to use the Mean Value Theorem, and we're trying to show something about

𝑒 𝑥

, so let's apply the MVT to

𝑓 (𝑥) = 𝑒 𝑥

on the interval

(𝑎, 𝑏)

and see what happens. Note that

𝑒 𝑥

is continuous and differentiable for all real

𝑥

, so the MVT applies.

𝑓 (𝑏) - 𝑓 (𝑎) 𝑏 - 𝑎 = 𝑓' (𝑐) 𝑒 𝑏 - 𝑒 𝑎 𝑏 - 𝑎 = 𝑓' (𝑐) 𝑒 𝑏 - 𝑒 𝑎 = 𝑓' (𝑐) (𝑏 - 𝑎)

Hmmm. We're a little stuck there. But looking back at the question, we see that we're after an inequality that involves

𝑒 𝑥

, so let's set

𝑏 = 𝑥

𝑓 (𝑏) - 𝑓 (𝑎) 𝑏 - 𝑎 𝑒 𝑥 - 𝑒 𝑎 = 𝑓' (𝑐) (𝑥 - 𝑎)

That's looking more promising, but we need to get rid of the

𝑎

. For lack of other ideas, let's try

𝑎 = 0

to make it go away. And note that

𝑒 0 = 1

𝑓 (𝑏) - 𝑓 (𝑎) 𝑏 - 𝑎 𝑒 𝑥 - 1 = 𝑓' (𝑐) (𝑥 - 0) = 𝑓' (𝑐) 𝑥 𝑒 𝑥 = 𝑓' (𝑐) 𝑥 + 1

Oh, that's now close! Let's think about

𝑓' (𝑐)

: we know

𝑓' (𝑥) = 𝑒 𝑥

, so

𝑓' (𝑐) = 𝑒 𝑐

𝑓 (𝑏) - 𝑓 (𝑎) 𝑏 - 𝑎 𝑒 𝑥 = 𝑒 𝑐 𝑥 + 1

But we're only interested in

𝑥 > 0

. And for

𝑐 > 0

, we know

𝑒 𝑐 > 1

. Hence

𝑒 𝑐 𝑥 > 𝑥

, and so

𝑓 (𝑏) - 𝑓 (𝑎) 𝑏 - 𝑎 𝑒 𝑥 > 𝑥 + 1 ✓

Whew, done! This "proof problem" is a good example of a question where we can't immediately see how it's going to turn out when we begin. Instead, you just have to dive in and start, and then try different things to keep moving forward, with the hope you'll land in the right place at the end. In this case, we had to just invoke that

𝑎 = 0

because we didn't want an

𝑎

in our final expression, but we didn't know that until we hit a certain point along the way. The crucial take away: Start, and then keep going. If one path doesn't work out, try another!

[close solution]

Practice Problem #9: Prove that only one root (based on an actual university exam question)

[Note: We examined a question similar to this one question in the topic "Curve Sketching," but used a different approach there—one that many students prefer because it seems more intuitive. Mathematicians, however, tend to prefer using Rolle's theorem as we illustrate here. You should become familiar with both approaches, and you might ask your instructor for any preference before an exam.]

Show that

𝑓 (𝑥) = 𝑥 5 + 6 𝑥 3 + 𝑥 - 4

has exactly one real root.

There are two parts to this proof:
(I) Show that the function has at least one root. We'll use the Intermediate Value Theorem for this part and show there exists some value

𝑐 1

(- \infty < 𝑐 1 < \infty)

for which

𝑓 (𝑐 1) = 0

.
(II) We'll then use Rolle's Theorem to show, by contradiction, that

𝑓 (𝑥)

does not have two or more roots.
Those two facts together mean that

𝑓 (𝑥)

has one and only one root.

(I) Let's first show that the function has one real root.
First note that

l i m 𝑥 \to - \infty 𝑓 (𝑥) = - \infty

, while

l i m 𝑥 \to \infty 𝑓 (𝑥) = \infty

. Since

𝑓

is a continuous function, by the Intermediate Value Theorem there must be some value

𝑐 1

(- \infty < 𝑐 1 < \infty)

for which

𝑓 (𝑐 1) = 0

. That is,

𝑓 (𝑥)

has at least one root.

(II) A proof by contradiction has three key steps:
1. Assume the opposite of what you want to prove.
2. Use that assumption to develop a new condition.
3. Show that the new condition contradicts your original assumption. Therefore your original assumption must be wrong, and so you've proved what you were originally after.

We'll proceed using those steps:
1. Let's assume that

𝑓 (𝑥)

has two real roots. We'll call them

𝑎

and

𝑏

, such that

𝑓 (𝑎) = 𝑓 (𝑏) = 0

.

2. Rolle's Theorem applies to

𝑓 (𝑥)

, since

𝑓

is a polynomial and is therefore continuous and differentiable everywhere. According to the Theorem, then, there must be a value

𝑐 2

on the interval

(𝑎, 𝑏)

such that:

𝑓' (𝑐 2) = 0

3. Let's compute

𝑓' (𝑥)

𝑓' (𝑥) = 𝑑 𝑑 𝑥 [𝑥 5 + 6 𝑥 3 + 𝑥 - 4] = 5 𝑥 4 + 18 𝑥 2 + 1

But

𝑥 4 \geq 0

and

𝑥 2 \geq 0

, so

𝑓' (𝑥) \geq 0 + 0 + 1 \geq 1 f o r a l l 𝑥

That is,

𝑓' (𝑥)

can never be 0. But in Step 2 above we saw that if there are two roots such that

𝑓 (𝑎) = 𝑓 (𝑏) = 0

, then there must be some value

𝑐 2

such that

𝑓' (𝑐 2) = 0

. So we have on the one hand

𝑓' (𝑥)

can never be 0, while on the other hand

𝑓' (𝑐 2) = 0

for some value of

𝑐 2

. This is a contradiction, and so our original assumption in Step 1 must have been incorrect:

𝑓

cannot have two real roots. (By extension,

𝑓

cannot have three or more roots either.)

Hence we have shown that (I)

𝑓

has at least one real root, and (II) it does not have two or more real roots. Therefore

𝑓

has exactly one real root.

✓

[close solution]

Practice Problem #10: Prove that only one root (based on an actual university exam question)

Show that

𝑔 (𝑥) = 6 𝑥 - c o s 𝑥

has exactly one real root.

We'll take the same approach we did in the preceding problem:
(I) show that the function has at least one root, and then
(II) use Rolle's Theorem and proof by contradiction to show that it does not have two or more roots.

(I) To show that

𝑔 (𝑥)

has at least one root, let's again use the Intermediate Value Theorem (IVT). To keep things simple, let's try some values for

𝑥

for which

c o s 𝑥 = 0

𝑥 = - 𝜋 / 2

and

𝜋 / 2

𝑔 (- 𝜋 / 2) = 6 (- 𝜋 2) - c o s (𝜋 2) = - 3 𝜋, w h i c h i s < 0 𝑔 (𝜋 / 2) = 6 (𝜋 2) - c o s (𝜋 2) = 3 𝜋, w h i c h i s > 0

Since

𝑔 (𝑥)

is a continuous function, we then know from the IVT that there exists

𝑐 1

- 𝜋 2 < 𝑐 1 < 𝜋 2

, such that

𝑔 (𝑐 1) = 0

. Thus

𝑔 (𝑥)

has at least one real root.

(II) We again use proof by contradiction, following the same three steps as we did in part (a):
1. Assume the opposite of what you want to prove.
2. Use that assumption to develop a new condition.
3. Show that the new condition contradicts your original assumption. Therefore your original assumption must be wrong, and so you've proved what you were originally after.

We'll proceed using those steps:
1. Let's assume that

𝑔 (𝑥)

has two real roots. We'll call them

𝑎

and

𝑏

, such that

𝑔 (𝑎) = 𝑔 (𝑏) = 0

.

2. Rolle's Theorem applies to

𝑔 (𝑥)

, since

𝑔

is the sum of a polynomial and the cosine function and is therefore continuous and differentiable everywhere. According to the Theorem, then, there must be a value

𝑐 2

on the interval

(𝑎, 𝑏)

such that:

𝑔' (𝑐 2) = 0

3. Let's compute

𝑔' (𝑥)

𝑔' (𝑥) = 𝑑 𝑑 𝑥 [6 𝑥 - c o s 𝑥] = 6 + s i n 𝑥 \geq 5

since the most-negative

s i n 𝑥

can be is

- 1 .

Hence

𝑔' (𝑥)

can never be zero. But in Step 2 above we saw that if there are two roots such that

𝑔 (𝑎) = 𝑔 (𝑏) = 0

, then there must be some value

𝑐 2

such that

𝑔 (𝑐 2) = 0

. So we have on the one hand

𝑔' (𝑥)

can never be 0, while on the other hand

𝑔' (𝑐 2) = 0

for some value of

𝑐 2

. This is a contradiction, and so our original assumption in Step 1 must have been incorrect:

𝑔

cannot have two real roots. (By extension,

𝑔

cannot have three or more roots either.)

Hence we have shown that (I)

𝑔

has at least one real root, and (II) it does not have two or more real roots. Therefore

𝑔

has exactly one real root.

✓

[close solution]

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Mean Value Theorem

Mean Value Theorem

Rolle's Theorem

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