Mean Value Theorem

Statement of the Theorem

If $f$ is:

1. Continuous on the closed interval $[a, b]$
2. Differentiable on the open interval $(a, b)$

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

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

Geometric Interpretation

The slope of the tangent line at some point $c$ equals the slope of the secant line connecting $(a, f(a))$ and $(b, f(b))$.

In other words: At some point, the instantaneous rate of change equals the average rate of change.

Related Theorems

Rolle's Theorem (Special Case)

If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists $c$ in $(a, b)$ where:

$f'(c) = 0$

Cauchy's Mean Value Theorem (Generalization)

If $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$, with $g'(x) \neq 0$, then there exists $c$ in $(a, b)$ such that:

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

Consequences

Constant Functions

If $f'(x) = 0$ for all $x$ in an interval, then $f$ is constant on that interval.

Increasing/Decreasing

• If $f'(x) > 0$ on $(a, b)$, then $f$ is increasing on $[a, b]$
• If $f'(x) < 0$ on $(a, b)$, then $f$ is decreasing on $[a, b]$

Equal Derivatives

If $f'(x) = g'(x)$ for all $x$ in an interval, then $f(x) = g(x) + C$ for some constant $C$.

Examples

Example 1: Finding c

For $f(x) = x^2$ on $[1, 3]$, find $c$ guaranteed by MVT.

$f'(x) = 2x$

$f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4$

$2c = 4$

$c = 2$

So at $x = 2$, the tangent slope equals the secant slope.

Example 2: Rolle's Theorem

For $f(x) = x^3 - 3x$ on $[-\sqrt{3}, \sqrt{3}]$, verify Rolle's and find $c$.

Check: $f(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0$ ✓ $f(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0$ ✓

$f'(x) = 3x^2 - 3 = 0$

$x^2 = 1$

$x = \pm 1$

Both $c = 1$ and $c = -1$ satisfy Rolle's Theorem.

Example 3: Proving an Inequality

Prove that $\sin(x) < x$ for $x > 0$.

Let $f(x) = x - \sin(x)$ on $[0, x]$.

$f(0) = 0$, $f'(t) = 1 - \cos(t) \geq 0$ for all $t$.

By MVT, for some $c$ in $(0, x)$:

$f(x) - f(0) = f'(c)(x - 0)$

$f(x) = f'(c) \cdot x \geq 0$

Since $f'(c) \geq 0$ and $x > 0$, we have $f(x) \geq 0$, so $x \geq \sin(x)$.

Moreover, $f'(t) = 0$ only at isolated points, so $f(x) > 0$ for $x > 0$.

Example 4: Speed Limit

A car travels 150 miles in 2 hours. Prove the car exceeded 70 mph at some point.

Let $s(t)$ = position at time $t$.

By MVT:

$s'(c) = \frac{s(2) - s(0)}{2 - 0} = \frac{150}{2} = 75 \text{ mph}$

At some time $c$, the instantaneous speed was exactly 75 mph > 70 mph.

Example 5: Uniqueness of Roots

Show $f(x) = x^3 + x - 1$ has exactly one real root.

$f(0) = -1 < 0$, $f(1) = 1 > 0$, so by IVT, at least one root exists in $(0, 1)$.

$f'(x) = 3x^2 + 1 > 0$ for all $x$, so $f$ is strictly increasing.

By MVT, if $f(a) = f(b) = 0$ with $a < b$, then $f'(c) = 0$ for some $c$, contradiction!

Therefore, exactly one root exists.