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3 years ago

If m,n ,then the value of x∧m/x∧n is _____

Mathematics

1 answer:

strojnjashka [21]3 years ago

7 0

Answer:

Step-by-step explanation:

$\dfrac{x^{m}}{x^{n}}=x^{m-n}$

In exponent division, if bases are same, then subtract the powers

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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 inte

MariettaO [177]

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.

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,

$f'(x)>0\\\Rightarrow cos(x)$

And f (x) is decreasing for,

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

(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,

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

And,

$f''(x)$

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

7 0

3 years ago