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gogolik [260]
3 years ago
14

Consider the function on the interval (0, 2π). f(x) = sin(x) cos(x) + 8 (a) Find the open interval(s) on which the function is i

ncreasing 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:
makkiz [27]3 years ago
7 0

Answer:

a) Increasing in

(0,\frac{\pi}{4})

(\frac{5\pi}{4},2\pi)

decreasing

(\frac{\pi}{4},\frac{5\pi}{4})

Local maximum

\frac{\pi}{4}

Local minimum

\frac{5\pi}{4}

Step-by-step explanation:

Let f(x) be

f(x) = sin(x)+cos(x)+0 for 0<x<2π.

Taking the first derivative

f'(x) = cos(x)-sin(x)

The critical points are those where the derivative vanishes.

f'(x) = 0 iif cos(x) = sin (x), so, the critical points in (0, 2π) are

x=\frac{\pi}{4}\;and\; \frac{5\pi}{4}

To find out what kind of critical points they are, we take the second derivative

f''(x) = -sin(x)-cos(x)

Evaluate this expression at the critical points

f''(\frac{\pi}{4})=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}< 0

so, this point is a local maximum.

f''(\frac{5\pi}{4})=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}> 0

and here we have a local minimum.

The function then is increasing in the intervals

(0,\frac{\pi}{4})\;and\;(\frac{5\pi}{4},2\pi)

and decreasing in

(\frac{\pi}{4},\frac{5\pi}{4})

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