Q. 60 - 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.

$f\left(x\right)=\tan \left(x\right), [a,b] = [-\mathrm{\pi },\mathrm{\pi }]$.

Step-by-Step Solution

Answer

The function $f\left(x\right)=\tan \left(x\right)$ satisfies the Mean Value Theorem and the value is $c=se{c}^{-1}\left(0\right)$.

Step 1. Given Information.

The function is,

$f\left(x\right)=\tan \left(x\right), [a,b]=[-\mathrm{\pi },\mathrm{\pi }]$.

Step 2. Mean Value Theorem.

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

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

$$\begin{gathered} \frac{f\left(\mathrm{\pi }\right)-\mathrm{f}(-\mathrm{\pi })}{\mathrm{\pi }-(-\mathrm{\pi })}=\frac{\tan \left(\mathrm{\pi }\right)-\tan \left(-\mathrm{\pi }\right)}{\mathrm{\pi }+\mathrm{\pi }} \\ =\frac{0-0}{2\mathrm{\pi }} \\ =\frac{0}{2\mathrm{\pi }} \\ =0 \end{gathered}$$

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

We have to find the value of $c$ with $f^{\text{'}}\left(c\right)=0$ we solve it:

$$\begin{gathered} se{c}^2\left(x\right)=0 \\ \Rightarrow sec\left(x\right)=0 \\ \Rightarrow x=se{c}^{-1}\left(0\right) \end{gathered}$$