What is the extreme f(x)=4cos(2x)+1,(0,2pi) ?

The extreme f(x)=4cos(2x)+1,(0,2pi) is Minimum(pi/2 ,-3),Maximum(pi,5),Minimum((3pi)/2 ,-3)

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extreme f(x)=4cos(2x)+1,(0,2pi)

extreme points

f (x) = 4 cos (2 x) + 1, (0, 2 π)

Minimum (\frac{π}{2}, - 3), Maximum (π, 5), Minimum (\frac{3 π}{2}, - 3)

Solution steps

Find the critical points:

x = \frac{π}{2}, x = π, x = \frac{3 π}{2}

Domain of

4 cos (2 x) + 1, (0, 2 π) : 0 < x < 2 π

Find intervals:

Decreasing

: 0 < x < \frac{π}{2},

Increasing

: \frac{π}{2} < x < π,

Decreasing

: π < x < \frac{3 π}{2},

Increasing

: \frac{3 π}{2} < x < 2 π

Plug the extreme point

x = \frac{π}{2}

into

4 cos (2 x) + 1 \Rightarrow y = - 3

Minimum (\frac{π}{2}, - 3)

Plug the extreme point

x = π

into

4 cos (2 x) + 1 \Rightarrow y = 5

Maximum (π, 5)

Plug the extreme point

x = \frac{3 π}{2}

into

4 cos (2 x) + 1 \Rightarrow y = - 3

Minimum (\frac{3 π}{2}, - 3)

Minimum (\frac{π}{2}, - 3), Maximum (π, 5), Minimum (\frac{3 π}{2}, - 3)

2 x y - 5 x^{2} - 2 y^{2} + 4 x + 4 y - 4

f (x) = 9 x e^{- x}

f (x) = - x^{3} + 12 x - 7

f (x) = x^{2} - 11, - 2 \leq x \leq 4

f (x) = 3 x^{3} - 36 x^{2} + 108 x + 8

What is the extreme f(x)=4cos(2x)+1,(0,2pi) ?
The extreme f(x)=4cos(2x)+1,(0,2pi) is Minimum(pi/2 ,-3),Maximum(pi,5),Minimum((3pi)/2 ,-3)