# Answer to Question #821 in Calculus for Max

Question #821

Where does the periodic function f(x) = 2e

Show all working and explain your answer clearly.

^{(sin x/2)}take on its extreme values andwhat are these values?&Show all working and explain your answer clearly.

Expert's answer

To find extreme points we have to solve the equation:

f'(x) = 0.

f'(x) =( 2e

As e

sin (x) = 0;

x = Pi n , n = Z (integers).

f'(x) = 0.

f'(x) =( 2e

^{sin(x/2)})'= 2 sin (x/2) e^{sin(x/2)}* 1/2 cos (x/2) = 1/2 sin(x) e^{sin(x/2) }= 0;As e

^{sin(x/2) }is always greater than 0, we can solve the equation only for sin(x) :sin (x) = 0;

x = Pi n , n = Z (integers).

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