📂Analysis

Mean Value Theorem in Analysis

Theorem

Let the functions $f$ and $g$ be continuous on the interval $[a,b]$ and differentiable on $(a,b)$. Then there exists $x \in (a,b)$ that satisfies the following equation.

$$[f(b)-f(a)]g^{\prime}(x)=[g(b)-g(a)]f^{\prime}(x)$$

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Note that differentiability is not necessary at the endpoints $a$ and $b$.

Explanation

This is a generalization of the Mean Value Theorem learned in high school and in calculus. If we set it as $g(x)=x$, it becomes the familiar form.

Corollary: Mean Value Theorem

Let the function $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ as a real function. Then there exists $x\in (a,b)$ that satisfies the following equation.

$$f(b)-f(a)=(b-a)f^{\prime}(x)$$

Proof

Let us define function $h$ as follows.

$$h(t) = [f(b)-f(a)]g(t) -[g(b)-g(a)]f(t)\quad (a\le t \le b)$$

Then $h$ is continuous on $[a,b]$ as the sum of continuous functions. Also, it is differentiable on $(a,b)$ as the sum of differentiable functions. Then the following holds.

$$h^{\prime}(t)= [f(b)-f(a)]g^{\prime}(t) -[g(b)-g(a)]f^{\prime}(t)$$

Also, the following equation holds.

$$\begin{equation} \begin{aligned} h(a) &= [f(b)-f(a)]g(a)-[g(b)-g(a)]f(a) \\ &= f(b)g(a)-f(a)g(b) \\ &= [f(b)-f(a)]g(b)-[g(b)-g(a)]f(b) \\ &= h(b) \end{aligned} \label{eq1} \end{equation}$$

Now, proving that for some $x\in (a,b)$, $h^{\prime}(x)=0$ holds concludes the proof.

• Case 1. If $h$ is constant

  For all $x \in (a,b)$, $h^{\prime}(x)=0$ holds.

• Case 2. If for some $t\in (a,b)$, $h(t) > h(a)$ holds

  By the Extreme Value Theorem, there exists a $x\in [a,b]$ where the function value of $h$ is maximum. Then by $\eqref{eq1}$, $x \in (a,b)$ holds. Then, $x$ is a local maximum of $h$ and differentiable at $x$, so $h^{\prime}(x)=0$ holds.

• Case 3. If for some $t \in (a,b)$, $h(t)<h(a)$ holds

  This case is similar to Case 2. where there exists a local minimum $x\in (a,b)$ of $h$, and $h^{\prime}(x)=0$ holds.

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