What is the extreme f(x)=4cos^2(x),0<= x<= pi ?

The extreme f(x)=4cos^2(x),0<= x<= pi is Maximum(0,4),Minimum(pi/2 ,0),Maximum(pi,4)

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extreme f(x)=4cos^2(x),0<= x<= pi

extreme points

f (x) = 4 cos^{2} (x), 0 \leq x \leq π

Maximum (0, 4), Minimum (\frac{π}{2}, 0), Maximum (π, 4)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

4 cos^{2} (x), 0 \leq x \leq π : 0 \leq x \leq π

Find intervals:

Decreasing

: 0 < x < \frac{π}{2},

Increasing

: \frac{π}{2} < x < π

Plug the extreme point

x = 0

into

4 cos^{2} (x) \Rightarrow y = 4

Maximum (0, 4)

Plug the extreme point

x = \frac{π}{2}

into

4 cos^{2} (x) \Rightarrow y = 0

Minimum (\frac{π}{2}, 0)

Plug the extreme point

x = π

into

4 cos^{2} (x) \Rightarrow y = 4

Maximum (π, 4)

Maximum (0, 4), Minimum (\frac{π}{2}, 0), Maximum (π, 4)

cos (t)

f (x) = x^{4} - 32 x^{2} + 8

f (x, y) = 2 x^{3} - 5 x y^{2} + 4 y^{3}

f (x, y) = 2 x^{2} - x y + y^{2} - 6 x - 9 y + 7

12 x^{\frac{2}{3}} - 8 x

What is the extreme f(x)=4cos^2(x),0<= x<= pi ?
The extreme f(x)=4cos^2(x),0<= x<= pi is Maximum(0,4),Minimum(pi/2 ,0),Maximum(pi,4)