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.
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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See Also
James Stewart, Daniel Clegg, and Saleem Watson, Calculus (early transcendentals, 9E), p291-292 ↩︎