Question

onsider the equation below. (If an answer does not exist, enter DNE.) f(x) = 8 cos2(x) − 16 sin(x), 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) (No Response) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (No Response) (b) Find the local minimum and maximum values of f. local minimum value (No Response) local maximum value (No Response) (c) Find the inflection points. (x, y) = (No Response) (smaller x-value) (x, y) = (No Response) (larger x-value)

Answer 1

Answer:

(a) Increasing:$\frac{\pi}{2}< x< \frac{3\pi}{2}$ and Decreasing:$0< x< \frac{\pi}{2}\ \text{or}\ \frac{3\pi}{2}< x< 2\pi$

(b) The local minimum and maximum values are -16 and 16 respectively.

(c) The inflection points are $(\frac{\pi}{6},\ -2)\ \text{and}\ (\frac{5\pi}{6},\ -2)$

Step-by-step explanation:

The function provided is:

$f(x)=8cos^{2}(x)-16sin( x);\ 0\leq x\leq 2\pi$

(a)

$f(x)=8cos^{2}(x)-16sin( x);\ 0\leq x\leq 2\pi$

Then, $f'(x)=-16cos(x)sin(x)-16cos(x)=-16cos(x)[1+sin(x)]$

Note, $1+sin(x)\geq 0\ \text{and }\ sin(x)\geq 1$

Then, $sin(x)=-1\Rightarrow x=\frac{3\pi}{2}$ for $0\leq x\leq 2\pi$.

Also $cos(x)=0$.

Thus, f (x) is increasing for,

$f'(x)>0\\\Rightarrow cos(x)<0\\\Rightarrow \frac{\pi}{2}< x< \frac{3\pi}{2}$

And f (x) is decreasing for,

$f'(x)<0\\\Rightarrow cos(x)>0\\\Rightarrow 0< x< \frac{\pi}{2}\ \text{or}\ \frac{3\pi}{2}< x< 2\pi$

(b)

From part (a) f (x) changes from decreasing to increasing at $x=\frac{\pi}{2}$ and from increasing  to decreasing at $x=\frac{3\pi}{2}$.

The local minimum value is:

$f(\frac{\pi}{2})=8cos^{2}(\frac{\pi}{2})-16sin(\frac{\pi}{2})=-16$

The local maximum value is:

$f(\frac{3\pi}{2})=8cos^{2}(\frac{3\pi}{2})-16sin(\frac{3\pi}{2})=16$

(c)

Compute the value of f'' (x) as follows:

$f''(x)=16sin(x)[1+sin(x)]-16cos^{2}(x)\\\\=16sin(x)+16sin^{2}(x)-16[1-sin^{2}(x)]\\\\=32sin^{2}(x)+16sin(x)-16\\\\=16[2sin(x)-1][sin (x)+1]$

So,

$f''(x)>0\\\Rightarrow sin(x)>\frac{1}{2}\\\Rightarrow \frac{\pi}{6}<x<\frac{5\pi}{6}$

And,

$f''(x)<0\\\\\Rightarrow sin(x)<\frac{1}{2}\ \text{and}\ sin (x)\neq -1\\\\\Rightarrow 0<x<\frac{\pi}{6}\ \text{or} \frac{5\pi}{6}<x<\frac{3\pi}{2}\ \text{or}\ \frac{3\pi}{2}<x<2\pi$

Thus, f (x) is concave upward on $(\frac{\pi}{6},\ \frac{5\pi}{6})$ and concave downward on $(0,\ \frac{\pi}{6}), (\frac{5\pi}{6},\ \frac{3\pi}{2})\ \text{and}\ (\frac{3\pi}{2},\ 2\pi)$.

If $x=\frac{\pi}{6}$, then f (x) will be:

$f(\frac{\pi}{6})=8cos^{2}(\frac{\pi}{6})-16sin(\frac{\pi}{6})=-2$

If $x=\frac{5\pi}{6}$, then f (x) will be:

$f(\frac{5\pi}{6})=8cos^{2}(\frac{5\pi}{6})-16sin(\frac{5\pi}{6})=-2$

The inflection points are $(\frac{\pi}{6},\ -2)\ \text{and}\ (\frac{5\pi}{6},\ -2)$.