The Mean Value Theorem for Derivatives

The Mean Value Theorem says: if the average rate of change of a function over an interval is $k$ , then the instantaneous rate of change must equal $k$ at least once inside that interval. In other words, the speedometer must read the average speed at some moment during any trip.

Condition	Conclusion
$f$ continuous on $[a, b]$
$f$ differentiable on $(a, b)$	There exists $c \in (a, b)$ with f'(c)=\dfrac{f(b)-f(a)}{b-a}

Geometric Motivation

Consider the graph of $y = f (x)$ and two points $A (a, f (a))$ and $B (b, f (b))$ . The slope of the chord $A B$ is:

\tan (∠ G A B) = \frac{f (b) - f (a)}{b - a} .

There is at least one point $P$ on the curve between $A$ and $B$ where the tangent is parallel to the chord $A B$ (slope equal to the slope of $A B$ ). If the $x$ -value of $P$ is $c$ , then:

f'(c)=\frac{f(b)-f(a)}{b-a}.

Graph of y = f(x) with secant line through A and B, and a tangent line at interior point P parallel to the secant — Geometric interpretation of the Mean Value Theorem: the tangent at $P$ is parallel to the secant $A B$ .

Thinking of $f (t)$ as the position of a particle, the left-hand side of the equation represents the average velocity over $[a, b]$ , and the right-hand side is the instantaneous velocity at some time $c$ . The theorem states that at some instant the instantaneous velocity must equal the average velocity.

The Theorem

The Mean Value Theorem for Derivatives. Let $f$ be a function with the following properties:

$f$ is continuous on a closed interval $[a, b]$ , and
$f$ is differentiable on the open interval $(a, b)$ .

Then there exists at least one point $c$ in the open interval $(a, b)$ such that:

f'(c)=\frac{f(b)-f(a)}{b-a},\tag{iii}

or equivalently,

f(b)-f(a)=f'(c)(b-a).\tag{iv}

Proof

The equation of the line through

A (a, f (a))

and

B (b, f (b))

is:

y = f (a) + \frac{f (b) - f (a)}{b - a} (x - a) .

Define a new function

h (x)

by subtracting this line from

f (x)

h (x) = f (x) - [f (a) + \frac{f (b) - f (a)}{b - a} (x - a)] .

Then

h

satisfies the hypotheses of Rolle's Theorem on

[a, b]

because: 1.

h

is continuous on

[a, b]

(as a difference of continuous functions), 2.

h

is differentiable on

(a, b)

, 3.

h (a) = h (b) = 0

. Therefore, there exists at least one

c \in (a, b)

such that h'(c)=0. Since: h'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}, setting h'(c)=0 gives: f'(c)=\frac{f(b)-f(a)}{b-a}.

Graph showing the auxiliary function h(x) = f(x) minus the secant line, satisfying Rolle's Theorem — The auxiliary function $h (x)$ satisfies Rolle's Theorem.

Remarks:

The Mean Value Theorem for derivatives is also called Lagrange's Mean Value Theorem.
The theorem does not locate $c$ , it only asserts existence. In practice, finding the exact value of $c$ is often difficult, but the mere existence of such a point allows us to deduce many important properties.
There may be more than one point satisfying the theorem.
If f' fails to exist even at one interior point, the theorem may not hold.

Graph showing two interior points where the tangent is parallel to the secant, illustrating that c need not be unique — There may be more than one point $c$ that satisfies the Mean Value Theorem.

The number $c$ can also be written as $c = a + θ (b - a)$ where $0 < θ < 1$ . Replacing $a$ by $x$ and $b$ by $x + h$ , the Mean Value Theorem can be written:

f(x+h)=f(x)+hf'(x+\theta h)\qquad\text{where }0<\theta<1.

Examples

Verify the conclusion of the Mean Value Theorem for $f (x) = x^{2} - 4 x + 3$ on $[- 2, 3]$ .

Solution

Since

f

is a polynomial, it is continuous and differentiable everywhere, so the hypotheses of the Mean Value Theorem are satisfied. The average rate of change over

[- 2, 3]

is:

\frac{f (3) - f (- 2)}{3 - (- 2)} = \frac{0 - 15}{5} = - 3.

The instantaneous rate of change is: f'(x)=2x-4. The Mean Value Theorem guarantees some

c \in (- 2, 3)

with f'(c)=-3:

2 c - 4 = - 3 \Rightarrow c = \frac{1}{2} .

Since

1 / 2 \in (- 2, 3)

, the value

c = 1 / 2

satisfies the conclusion.

Graph of f(x) = x^2 - 4x + 3 on [-2,3] with the secant and the tangent at c = 1/2 drawn

Show that:

(a) If -1\leq f'(x)\leq1 for every $x$ , then $| f (x) - f (y) | \leq | x - y |$ for every $x$ and $y$ .

(b) For every $x$ and $y$ : $| \sin x - \sin y | \leq | x - y |$ .

Solution

**(a)** Since f'(x) exists,

f

is differentiable and continuous everywhere. By the Mean Value Theorem, there exists some

c

between

x

and

y

such that: f(x)-f(y)=f'(c)(x-y). Taking absolute values: |f(x)-f(y)|=|f'(c)|\ |x-y|. Since -1\leq f'(x)\leq1 for every

x

, we have |f'(c)|\leq1, and thus:

| f (x) - f (y) | \leq | x - y | .

**(b)** Let

f (x) = \sin x

. Then: -1\leq f'(x)=\cos x\leq1. It follows from part (a) that:

| \sin x - \sin y | \leq | x - y | .

Show that if $x > 0$ , then $\sqrt{1 + x} < 1 + \frac{1}{2} x$ .

Solution

Let

f (x) = \sqrt{1 + x}

. Then f'(x)=\dfrac{1}{2\sqrt{1+x}}. The function

f

is continuous and differentiable for all

x > - 1

. Let

a = 0

and

b = x > 0

. By the Mean Value Theorem, there is a number

c

between $0$ and

x

such that: f(x)-f(0)=f'(c)(x-0) \sqrt{1+x}-1=f'(c)\cdot x.\tag{i} We claim f'(c)<\dfrac{1}{2}, because for

c > 0

: 1<\sqrt{1+c}\Rightarrow\frac{1}{\sqrt{1+c}}<1\Rightarrow f'(c)=\frac{1}{2\sqrt{1+c}}<\frac{1}{2}.\tag{ii} Combining (i) and (ii):

\sqrt{1 + x} - 1 < \frac{1}{2} x \Rightarrow \sqrt{1 + x} < 1 + \frac{1}{2} x .

Cauchy's Mean Value Formula

Cauchy's Mean Value Formula. Assume $f (x)$ and $g (x)$ are continuous on $[a, b]$ and differentiable on $(a, b)$ . Then there exists at least one point $c$ in $(a, b)$ such that:

f'(c)\left[g(b)-g(a)\right]=g'(c)\left[f(b)-f(a)\right].

If g'(c)\neq0 and $g (a) \neq g (b)$ , this is equivalent to:

\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}.

Proof

Define:

h (x) = f (x) [g (b) - g (a)] - g (x) [f (b) - f (a)] .

Since

f

and

g

are continuous on

[a, b]

and differentiable on

(a, b)

, so is

h

. One can verify that

h (a) = h (b) = f (a) g (b) - f (b) g (a)

. Therefore, Rolle's Theorem applies to

h

, and there exists

c \in (a, b)

with h'(c)=0: h'(c)=f'(c)\left[g(b)-g(a)\right]-g'(c)\left[f(b)-f(a)\right]=0, giving the result.

The Mean Value Theorem (Lagrange's theorem) is a special case of Cauchy's Mean Value Formula when $g (x) = x$ .

Frequently Asked Questions

What is the physical meaning of the Mean Value Theorem?

If you drive 300 km in 3 hours, your average speed is 100 km/h. The Mean Value Theorem guarantees that at some moment during the trip, your speedometer read exactly 100 km/h. More generally, the average rate of change over any interval must be achieved instantaneously at some interior point.

How is the Mean Value Theorem different from Rolle's Theorem?

Rolle's Theorem is the special case of the Mean Value Theorem where

f (a) = f (b)

. In that case, the average rate of change is

0 / something = 0

, so the conclusion becomes f'(c)=0 for some

c \in (a, b)

. In fact, the proof of the Mean Value Theorem uses Rolle's Theorem.

Can we find the exact value of c guaranteed by the theorem?

In many problems, yes, we solve f'(c)=(f(b)-f(a))/(b-a) for

c

. But the theorem guarantees existence even when finding

c

explicitly is impractical. Many important applications (such as establishing inequalities or proving uniqueness of solutions) only need to know that such a

c

exists.

Geometric Motivation

The Theorem

Examples

Cauchy's Mean Value Formula

Frequently Asked Questions

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