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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

2 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

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

Maximum (0, 2)

Plug the extreme point

x = \frac{π}{2}

into

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

Minimum (\frac{π}{2}, 0)

Plug the extreme point

x = π

into

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

Maximum (π, 2)

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

f (x, y) = x y + \frac{2}{x} + \frac{4}{y}

f (x, y) = - 5 x^{2} - 8 y^{2} - 2 x y + 42 x + 102 y

\frac{12}{x ^{2} + 1}

f (x) = cos (x) - 2 x, 0 \leq x \leq 4 π

f (x) = \frac{7 ( x - 1 ) ^{2}}{x + 9}

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