Question

2. Consider the function f(x) = sinx 2+cos a) What is the domain of the function? b) Use the first derivative to locate the intervals of increase or decrease as w e the intervals of increase or decrease as well as any local your analysis to the intervalo Sxs 21. Indicate the exact coordinates of the al extrema and clearly specify whether the point is a local max or local min. use the second derivative to determine intervals of concavity and any inflection points. Limit your analysis to the mit your analysis to the interval O Sxs 21. Indicate the exact coordinates of any inflection points. d) Sketch a graph of the function on the interval OS X SZT.

Answer 1

Given function is :

sinc f(3) = 2 + cos2

a) Domain of the function is =

(-2, 0)

b) Now, gives,

daft) = 0

(2 + cos.) * cos r - sin r * (-sinc) 2 (2 + cos.c)2 = 0

i.e.,

2 cosc + cos2x + sinac (2 + cos2

i.e., $2\cos x+1=0*(2+\cos x)^2$

i.e.,

2 cos .r+1=0

i.e.,

2 cos r = -1

i.e., $\cos x=-\frac{1}{2}$

i.e.,

I= cos -1

i.e.,

х = 27 4л 3 3

At ,

=

21 sin

i.e.,

( = ༡༤ (1/2) V/3/2

i.e., $f(\frac{2\pi}{3})=\frac{\sqrt{3}/2}{3/2}$

i.e., $f(\frac{2\pi}{3})=\frac{1}{\sqrt{3}}$

At $x=\frac{4\pi}{3}$ ,

sin SC ) = 2 + cos

i.e.,

101) = -V3/2 2-(1/2)

i.e., $f(\frac{4\pi}{3})=\frac{-\sqrt{3}/2}{3/2}$

i.e., $f(\frac{4\pi}{3})=-\frac{1}{\sqrt{3}}$

This implies that the function is increasing in the interval $\left ( 0,\frac{2\pi}{3} \right )\cup\left ( \frac{4\pi}{3},2\pi \right )$ and the function is decreasing in the interval $\left ( \frac{2\pi}{3},\frac{4\pi}{3} \right )$ .

Here, local max is $\left ( \frac{2\pi}{3},\frac{1}{\sqrt{3}} \right )$ and local min is $\left ( \frac{4\pi}{3},-\frac{1}{\sqrt{3}} \right )$ .

c) Now, $\frac{d^2}{dx^2}f(x)=0$ gives,

$\frac{d}{dx}\left (\frac{2\cos x+1}{(2+\cos x)^2} \right )=0$

i.e., $\frac{(2+\cos x)^2*(-2\sin x)-(2\cos x+1)*2(2+\cos x)(-\sin x)}{(2+\cos x)^4}=0$

i.e., $\frac{(2+\cos x)*(-2\sin x)+(2\cos x+1)*2(\sin x)}{(2+\cos x)^3}=0$

i.e., $\frac{-4\sin x-2\cos x\sin x+4\sin x\cos x+2\sin x}{(2+\cos x)^3}=0$

i.e., $-4\sin x-2\cos x\sin x+4\cos x\sin x+2\sin x=0$

i$2\cos x\sin x-2\sin x=0$

i.e., $\sin x(2\cos x-2)=0$

i.e., $\sin x(\cos x-1)=0$

i.e., $x=0,\ \pi, 2\pi$

Since x = 0 and $x=2\pi$ are the staring and ending points of the function, therefore $x=\pi$ is the inflection points of the function.

Now, $f''\left ( \frac{2\pi}{3} \right )=\frac{\sin \frac{2\pi}{3}\left ( \cos\frac{2\pi}{3}-1 \right )}{\left ( 2+\cos\frac{2\pi}{3} \right )^3}$

i.e., $f''\left ( \frac{2\pi}{3} \right )=\frac{\frac{\sqrt{3}}{2}\left ( -\frac{1}{2}-1 \right )}{\left ( 2-\frac{1}{2} \right )^3}$

i.e., $f''\left ( \frac{2\pi}{3} \right )=-\frac{\frac{3\sqrt{3}}{4}}{\left ( \frac{3}{2} \right )^3}$

i.e., $f''\left ( \frac{2\pi}{3} \right )<0$

Similarly we get, $f''\left ( \frac{4\pi}{3} \right )>0$ .

Therefore, the graph of the function is concave upward in the interval
(π, 2π)
and concave downward in the interval $(0,\pi)$.

d) Graph of the above function is :

(20/3, 0.577) (, 0) (41/3, -0.577)