If you look at $10 we are at 10 animals, but if you look at $100, we are at about 55 animals. Divide the two to find your average. That will give you the answer.

3 0

3 years ago

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Consider the function on the interval (0, 2π). f(x) = 7 sin2(x) + 7 sin(x) (a) Find the open interval(s) on which the function i

-BARSIC- [3]

Answer with Step-by-step explanation:

Given

$f(x)=7sin(2x)+7sin(x)$

Differentiating both sides by 'x' we get

$14cos(2x)+7cos(x)=f'(x)$

Now we know that for an increasing function we have

$f'(x)>0\\\\14cos(2x)+7cos(x)>0\\\\2cos(2x)+cos(x)>0\\\\2(2cos^{2}(x)-1)+cos(x)>0\\\\4cos^{2}(x)+cos(x)-2>0\\\\(2cos(x)+\frac{1}{2})^2-2-\frac{1}{4}>0\\\\(2cos(x)+\frac{1}{2})^2>\frac{9}{4}\\\\2cos(x)>\frac{3}{2}-\frac{1}{2}\\\\\therefore cos(x)>\frac{1}{4}\\\\\therefore x=[0,cos^{-1}(1/4)]\cup [2\pi-cos^{-1}(1/4),2\pi ]$

Similarly for decreasing function we have

$[tex]f'(x)$

Part b)

To find the extreme points we equate the derivative with 0

$f'(x)=0\\\\cos(x)=\frac{1}{4}\\\\x=cos^{-1}(\frac{1}{4})$

Thus point of extrema is only 1.

4 0

4 years ago