Q. 59 - Calculus

Question

Determine whether or not each function $f$ satisfies the hypotheses of the Mean Value Theorem on the given interval $[a,b]$. For those that do, use derivatives and algebra to find the exact values of all $c\in (a,b)$ that satisfy the conclusion of the Mean Value Theorem.

$f\left(x\right)=\sin \left(x\right), [a,b]=[0,\frac{\mathrm{\pi }}{2}]$.

Step-by-Step Solution

Answer

The function $f\left(x\right)=\sin \left(x\right)$ satisfies the Mean Value Theorem and the value is, $c={\cos }^{-1}\left(\frac{2}{\mathrm{\pi }}\right)$.

Step 1. Given Information.

The given function is,

$f\left(x\right)=\sin \left(x\right), [a,b]=[0,\frac{\mathrm{\pi }}{2}]$.

Step 2. Proving Mean Value Theorem.

The function $f\left(x\right)=\sin \left(x\right)$ is continuous and differentiable on $[0,\frac{\mathrm{\pi }}{2}]$. The Mean Value Theorem applies to this function on the interval $[0,\frac{\mathrm{\pi }}{2}]$.

The slope of the line from $(0,f(0\left)\right)$ to $(\frac{\mathrm{\pi }}{2},f(\frac{\mathrm{\pi }}{2}\left)\right)$ is:

$$\begin{gathered} \frac{f\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)-f\left(0\right)}{{\displaystyle \frac{\mathrm{\pi }}{2}}-0}=\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)-\sin \left(0\right)}{{\displaystyle \frac{\mathrm{\pi }}{2}}-0} \\ =\frac{1-0}{{\displaystyle \frac{\mathrm{\pi }}{2}}-0} \\ =\frac{1}{{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ =\frac{2}{\mathrm{\pi }} \end{gathered}$$

By the Mean Value Theorem, there must exist at least one point $c\in (0,\frac{\mathrm{\pi }}{2})$ with $f^{\text{'}}\left(c\right)=\frac{2}{\mathrm{\pi }}$

We have to find the value of $c$ with $f^{\text{'}}\left(c\right)=\frac{2}{\mathrm{\pi }}$ we solve it:

$$\begin{gathered} \cos \left(c\right)=\frac{2}{\mathrm{\pi }} \\ \Rightarrow c={\cos }^{-1}\left(\frac{2}{\mathrm{\pi }}\right) \end{gathered}$$