💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# (a) Find the intervals on which $f$ is increasing or decreasing.(b) Find the local maximum and minimum values of $f$.(c) Find the intervals of concavity and the inflection points.$f(x) = \sin x + \cos x$, $0 \leqslant x \leqslant 2\pi$

## (a) $f(x)=\sin x+\cos x, 0 \leq x \leq 2 \pi, \quad f^{\prime}(x)=\cos x-\sin x=0 \Rightarrow \cos x=\sin x \Rightarrow 1=\frac{\sin x}{\cos x} \Rightarrow$$\tan x=1 \Rightarrow x=\frac{\pi}{4} \text { or } \frac{\mathrm{sin}}{4} \text { . Thus, } f^{\prime}(x)>0 \Leftrightarrow \cos x-\sin x>0 \Leftrightarrow \cos x>\sin x \Leftrightarrow 0<x<\frac{\pi}{4} \text { or }$$\frac{5 \pi}{4}<x<2 \pi$ and $f^{\prime}(x)<0 \Leftrightarrow \cos x<\sin x \Leftrightarrow \frac{\pi}{4}<x<\frac{5 \pi}{4} .$ So $f$ is increasing on $\left(0, \frac{\pi}{4}\right)$ and $\left(\frac{5 \pi}{4}, 2 \pi\right)$ and $f$is decreasing on $\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$(b) $f$ changes from increasing to decreasing at $x=\frac{\pi}{4}$ and from decreasing to increasing at $x=\frac{5 \pi}{4} .$ Thus, $f\left(\frac{\pi}{4}\right)=\sqrt{2}$ is alocal maximum value and $f\left(\frac{5 \pi}{4}\right)=-\sqrt{2}$ is a local minimum value.(c) $f^{\prime \prime}(x)=-\sin x-\cos x=0 \Rightarrow-\sin x=\cos x \Rightarrow \tan x=-1 \Rightarrow x=\frac{3 \pi}{4}$ or $\frac{7 \pi}{4} .$ Divide the interval$(0,2 \pi)$ into subintervals with these numbers as endpoints and complete a second derivative chart.

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

Derivatives

Differentiation

Volume

Lectures

Join Bootcamp