Chapter 4.1, Problem 15E

To determine

To find : extreme values of the function on the interval using the Candidates test.

Answer to Problem 15E

Maximum occur at $\left(\frac{\pi }{4},1\right)$ and minimum occur at $\left(\frac{5\pi }{4},-1\right)$ .

Explanation of Solution

Given information :

  $f\left(x\right)=\sin \left(x+\frac{\pi }{4}\right)$ , $0\le x\le \frac{7\pi }{4}$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $f\left(x\right)=\sin \left(x+\frac{\pi }{4}\right)$ with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\sin \left(x+\frac{\pi }{4}\right)\\ =\cos \left(x+\frac{\pi }{4}\right)\cdot \frac{d}{dx}\left(x+\frac{\pi }{4}\right)\\ =\cos \left(x+\frac{\pi }{4}\right)\cdot 1\\ =\cos \left(x+\frac{\pi }{4}\right)\end{array}$

Take $\frac{dy}{dx}=0$ .

  $\begin{array}{c}\cos \left(x+\frac{\pi }{4}\right)=0\\ x+\frac{\pi }{4}={\cos }^{-1}\left(0\right)\\ x+\frac{\pi }{4}=\frac{\pi }{2}orx+\frac{\pi }{4}=\frac{3\pi }{2}\\ x=\frac{\pi }{2}-\frac{\pi }{4}orx=\frac{3\pi }{2}-\frac{\pi }{4}\\ x=\frac{\pi }{4}orx=\frac{5\pi }{4}\end{array}$

Two critical points exist in the given interval.

Plug the critical points in the function.

  $\begin{array}{c}f\left(\frac{\pi }{4}\right)=\sin \left(\frac{\pi }{4}+\frac{\pi }{4}\right)\\ =\sin \left(\frac{\pi }{2}\right)\\ =1\end{array}$

And

  $\begin{array}{c}f\left(\frac{5\pi }{4}\right)=\sin \left(\frac{5\pi }{4}+\frac{\pi }{4}\right)\\ =\sin \left(\frac{3\pi }{2}\right)\\ =-1\end{array}$

Plug the end points in the function.

  $\begin{array}{c}f\left(0\right)=\sin \left(0+\frac{\pi }{4}\right)\\ =\sin \left(\frac{\pi }{4}\right)\\ =\frac{1}{\sqrt{2}}\end{array}$

And

  $\begin{array}{c}f\left(\frac{7\pi }{4}\right)=\sin \left(\frac{7\pi }{4}+\frac{\pi }{4}\right)\\ =\sin \left(2\pi \right)\\ =0\end{array}$

Maximum occur at $\left(\frac{\pi }{4},1\right)$ and minimum occur at $\left(\frac{5\pi }{4},-1\right)$ .