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mr_godi [17]
4 years ago
13

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

s increasing or decreasing. (Enter your answers using interval notation.) increasing decreasing (b) Apply the First Derivative Test to identify all relative extrema. relative maxima (x, y) = (smaller x-value) (x, y) = (larger x-value) relative minima (x, y) = (smaller x-value) (x, y) = (larger x-value)
Mathematics
1 answer:
-BARSIC- [3]4 years ago
4 0

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.

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A

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Step-by-step explanation:

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