Mean Value Theorem (MVT)

Proof

Create the function $h(x)$ defined as:

(you can think of $h$ as the secant line between $a,b$). Notice that $h$ is differentiable by Algebraic Differentiability Theorem, where:

• $h(a)=f(a)$
• $h(b)=f(b)$
• $h^{\prime }(x)=\frac{f(b)-f(a)}{b-a}$ for all $x\in [a,b]$.

Then for $x\in [a,b]$ define:

Then $g(a)=0=g(b)$ so by Rolle's Theorem (Baby MVT) then $\mathrm{\exists }c\in (a,b)$ such that $g^{\prime }(c)=0$, namely:

as desired.

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What's cool here is that for:

on the RHS we only have a constant, but on the left hand side we have information about $f^{\prime }(c)$. Thinking, we could try different $a,b$ values to learn more information about $f^{\prime }$ at points within those intervals (however, you don't know the exact value of $c$ though).

Let's apply the MVT!

Examples

Let $f$ be differentiable on the interval $I$. If $f^{\prime }(x)=0$ for all $x\in I$ then $x$ is constant. We use this idea of anti-derivation all the time, but using the Mean Value Theorem (MVT) we get that for free.

Proof

Take any $x_1,x_2\in I$ and WLOG $x_1<x_2$. Now since $f$ is differentiable on $I$ then it satisfies the properties for Mean Value Theorem (MVT). As such then $\mathrm{\exists }c\in [x_1,x_2]$ such that:

But we know that $f^{\prime }(c)=0$ (per our given) so then:

so since $x_1,x_2$ were arbitrary then all points of the function must be equal, so $f$ is constant.

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See this again in Derivative is Zero Implies Constant Function.