Chapter 4.3, Problem 11E

(a)

To determine

To find:The interval on which $f$ is increasing.

Explanation of Solution

Given: The function is $f\left(x\right)=\sin x+\cos x$

Find the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\cos x-\sin x\\ 0=\cos x-\sin x\\ \sin x=\cos x\\ x=\frac{\pi }{4},\frac{5\pi }{4}\end{array}$

The function increases in the interval $\left(0,\frac{\pi }{4}\right)\cup \left(\frac{5\pi }{4},2\pi \right)$ .

The function decreases in the interval $\left(\frac{\pi }{4},\frac{5\pi }{4}\right)$

(b)

To determine

To find :The local maxima and minima.

Explanation of Solution

The function increases in the interval $\left(0,\frac{\pi }{4}\right)\cup \left(\frac{5\pi }{4},2\pi \right)$ .

The function decreases in the interval $\left(\frac{\pi }{4},\frac{5\pi }{4}\right)$

So, there is a local maxima at,

  $\left(\frac{\pi }{4},f\left(\frac{\pi }{4}\right)\right)=\left(\frac{\pi }{4},\sqrt{2}\right)$

The local minima is at,

  $\left(\frac{5\pi }{4},f\left(\frac{5\pi }{4}\right)\right)=\left(\frac{5\pi }{4},-\sqrt{2}\right)$

(c)

To determine

To find :The interval ofconcavity and points of inflection

Explanation of Solution

The graph is concave up when $f^{″}\left(x\right)>0$ and concave downward when $f^{″}\left(x\right)<0$ .

The derivative of $f^{\prime }\left(x\right)$ is positive when $f^{\prime }\left(x\right)$ is increasing.

The derivative of $f^{\prime }\left(x\right)$ is negative when $f^{\prime }\left(x\right)$ is decreasing

Calculate the second derivative.

  $\begin{array}{c}{f}^{″}\left(x\right)=-\sin x-\cos x\\ 0=-\sin x-\cos x\\ \tan x=-1\\ x=\frac{3\pi }{4},\frac{7\pi }{4}\end{array}$

The inflection point is $x=\frac{3\pi }{4},\frac{7\pi }{4}$ .

The function is concave up at $\left(\frac{3\pi }{4},\frac{7\pi }{4}\right)$ and concave down at $\left[0,\frac{3\pi }{4}\right)\cup \left(\frac{7\pi }{4},2\pi \right]$

Therefore, the function is concave upward at $\left(\frac{3\pi }{4},\frac{7\pi }{4}\right)$ , concave downward at $\left[0,\frac{3\pi }{4}\right)\cup \left(\frac{7\pi }{4},2\pi \right]$ and the point of inflection is $x=\frac{3\pi }{4},\frac{7\pi }{4}$ .