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

The extreme f(x)=cos^2(x),0<= x<= pi is Maximum(0,1),Minimum(pi/2 ,0),Maximum(pi,1)

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

extreme points

f (x) = cos^{2} (x), 0 \leq x \leq π

Maximum (0, 1), Minimum (\frac{π}{2}, 0), Maximum (π, 1)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

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

cos^{2} (x) \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = \frac{π}{2}

into

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

Minimum (\frac{π}{2}, 0)

Plug the extreme point

x = π

into

cos^{2} (x) \Rightarrow y = 1

Maximum (π, 1)

Maximum (0, 1), Minimum (\frac{π}{2}, 0), Maximum (π, 1)

f (x) = x^{3} + 2 x^{2} - 7 x

f (x, y) = x e^{- 3 x^{2} - 3 y^{2}}

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 2

\frac{1}{x - 2}

f (x) = - \frac{5 e ^{- x}}{5 - 3 x}

What is the extreme f(x)=cos^2(x),0<= x<= pi ?
The extreme f(x)=cos^2(x),0<= x<= pi is Maximum(0,1),Minimum(pi/2 ,0),Maximum(pi,1)