Proof of the Mean Value Theorem in Calculus 📂Calculus

Proof of the Mean Value Theorem in Calculus

Theorem¹

If the function $f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$ , then there exists at least one $c$ in $(a,b)$ that satisfies $\displaystyle f '(c)={{f(b)-f(a)}\over{b-a}}$ .

Description

It’s not just commonly used; it’s so famous that it’s abbreviated as MVT. The term ‘mean value’ comes from the idea that there is a point where the derivative equals the average rate of change over the entire interval. The concept of the average is so useful that there are various modified forms of it for application in many fields.

Proof

Let $\displaystyle m:= {{f(b)-f(a)}\over{b-a}}$ and define $g(x):=f(x)-mx$ , then $g(b)=g(a)$ and $g(x)$ is differentiable.

Rolle’s Theorem
If the function $f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$ , and if $f(a)=f(b)$ , then there exists at least one $c$ in $(a,b)$ that satisfies $f ' (c)=0$ .

By Rolle’s Theorem, there exists at least one $c$ in $(a,b)$ that satisfies $g ' (c)=0$ , and since $g ' (x)=f ' (x) - m$ , it follows that $g ' (c) = f '(c) - m = 0$ . By rearranging $f ' (c) -m = 0$ into $(-m)$ , we can find that there exists at least one $c$ in $(a,b)$ satisfying $\displaystyle f '(c) = m = {{f(b)-f(a)}\over{b-a}}$ .

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