What is the extreme f(x)=cos(x)+2cos^2(x) ?

The extreme f(x)=cos(x)+2cos^2(x) is

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extreme f(x)=cos(x)+2cos^2(x)

extreme points

f (x) = cos (x) + 2 cos^{2} (x)

Maximum (2 πn, 3), Minimum (arccos (- \frac{1}{4}) + 2 πn, - \frac{1}{8}), Maximum (π + 2 πn, 1), Minimum (- arccos (- \frac{1}{4}) + 2 π + 2 πn, - \frac{1}{8})

Solution steps

Find the critical points:

x = 2 πn, x = arccos (- \frac{1}{4}) + 2 πn, x = π + 2 πn, x = - arccos (- \frac{1}{4}) + 2 π + 2 πn

Domain of

cos (x) + 2 cos^{2} (x) : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn < x < arccos (- \frac{1}{4}) + 2 πn,

Increasing

: arccos (- \frac{1}{4}) + 2 πn < x < π + 2 πn,

Decreasing

: π + 2 πn < x < - arccos (- \frac{1}{4}) + 2 π + 2 πn,

Increasing

: - arccos (- \frac{1}{4}) + 2 π + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = 2 πn

into

cos (x) + 2 cos^{2} (x) \Rightarrow y = 3

Maximum (2 πn, 3)

Plug the extreme point

x = arccos (- \frac{1}{4}) + 2 πn

into

cos (x) + 2 cos^{2} (x) \Rightarrow y = - \frac{1}{8}

Minimum (arccos (- \frac{1}{4}) + 2 πn, - \frac{1}{8})

Plug the extreme point

x = π + 2 πn

into

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

Maximum (π + 2 πn, 1)

Plug the extreme point

x = - arccos (- \frac{1}{4}) + 2 π + 2 πn

into

cos (x) + 2 cos^{2} (x) \Rightarrow y = - \frac{1}{8}

Minimum (- arccos (- \frac{1}{4}) + 2 π + 2 πn, - \frac{1}{8})

Maximum (2 πn, 3), Minimum (arccos (- \frac{1}{4}) + 2 πn, - \frac{1}{8}), Maximum (π + 2 πn, 1), Minimum (- arccos (- \frac{1}{4}) + 2 π + 2 πn, - \frac{1}{8})

(2 y^{2} + x^{2}) e^{- (x^{2} + y^{2} - 3)}

\frac{x ^{5} - 5 x}{5}

f (x) = 12 x - 8

\frac{x ^{2} + 9}{2 x}

f (x) = x^{4} - 8 x^{2} + 16 [- 1.4]

What is the extreme f(x)=cos(x)+2cos^2(x) ?
The extreme f(x)=cos(x)+2cos^2(x) is