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

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

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extreme f(x)=2cos(2x)-2,(0,2pi)

extreme points

f (x) = 2 cos (2 x) - 2, (0, 2 π)

Minimum (\frac{π}{2}, - 4), Maximum (π, 0), Minimum (\frac{3 π}{2}, - 4)

Solution steps

Find the critical points:

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

Domain of

2 cos (2 x) - 2, (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

2 cos (2 x) - 2 \Rightarrow y = - 4

Minimum (\frac{π}{2}, - 4)

Plug the extreme point

x = π

into

2 cos (2 x) - 2 \Rightarrow y = 0

Maximum (π, 0)

Plug the extreme point

x = \frac{3 π}{2}

into

2 cos (2 x) - 2 \Rightarrow y = - 4

Minimum (\frac{3 π}{2}, - 4)

Minimum (\frac{π}{2}, - 4), Maximum (π, 0), Minimum (\frac{3 π}{2}, - 4)

y = x^{4} - 6 x^{2} = x^{2} (x^{2} - 6)

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

x^{4} - 16

f (x) = - x^{\frac{2}{3}} (x - 4)

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

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