What is the extreme f(x)=5cos^2(x),(0,pi) ?

The extreme f(x)=5cos^2(x),(0,pi) is Minimum(pi/2 ,0)

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extreme f(x)=5cos^2(x),(0,pi)

extreme points

f (x) = 5 cos^{2} (x), (0, π)

Minimum (\frac{π}{2}, 0)

Solution steps

Find the critical points:

x = \frac{π}{2}

Domain of

5 cos^{2} (x), (0, π) : 0 < x < π

Find intervals:

Decreasing

: 0 < x < \frac{π}{2},

Increasing

: \frac{π}{2} < x < π

Plug the extreme point

x = \frac{π}{2}

into

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

Minimum (\frac{π}{2}, 0)

f (x) = x^{2} (4 - x)^{2}

f (x, y) = 50 + 2 x + 3 y - 3 x^{2} - y^{2} - 0.5 x y

U (x, y) = - x^{2} - 3 y^{2} + 2 x y + 100 x

f (x) = x^{2} + \frac{120}{x}

f (x, y) = \frac{1}{2} y^{4} - 4 x y + 2 x^{4}

What is the extreme f(x)=5cos^2(x),(0,pi) ?
The extreme f(x)=5cos^2(x),(0,pi) is Minimum(pi/2 ,0)