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

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

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

extreme points

f (x) = sin (x) + cos (2 x)

Maximum (arcsin (\frac{1}{4}) + 2 πn, \frac{9}{8}), Minimum (\frac{π}{2} + 2 πn, 0), Maximum (π - arcsin (\frac{1}{4}) + 2 πn, \frac{9}{8}), Minimum (\frac{3 π}{2} + 2 πn, - 2)

Solution steps

Find the critical points:

x = arcsin (\frac{1}{4}) + 2 πn, x = \frac{π}{2} + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn, x = \frac{3 π}{2} + 2 πn

Domain of

sin (x) + cos (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < arcsin (\frac{1}{4}) + 2 πn,

Decreasing

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

Increasing

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

Decreasing

: π - arcsin (\frac{1}{4}) + 2 πn < x < \frac{3 π}{2} + 2 πn,

Increasing

: \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = arcsin (\frac{1}{4}) + 2 πn

into

sin (x) + cos (2 x) \Rightarrow y = \frac{9}{8}

Maximum (arcsin (\frac{1}{4}) + 2 πn, \frac{9}{8})

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

sin (x) + cos (2 x) \Rightarrow y = 0

Minimum (\frac{π}{2} + 2 πn, 0)

Plug the extreme point

x = π - arcsin (\frac{1}{4}) + 2 πn

into

sin (x) + cos (2 x) \Rightarrow y = \frac{9}{8}

Maximum (π - arcsin (\frac{1}{4}) + 2 πn, \frac{9}{8})

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

sin (x) + cos (2 x) \Rightarrow y = - 2

Minimum (\frac{3 π}{2} + 2 πn, - 2)

Maximum (arcsin (\frac{1}{4}) + 2 πn, \frac{9}{8}), Minimum (\frac{π}{2} + 2 πn, 0), Maximum (π - arcsin (\frac{1}{4}) + 2 πn, \frac{9}{8}), Minimum (\frac{3 π}{2} + 2 πn, - 2)

f (x) = \frac{1}{2}

\frac{x ^{2} + x - 2}{x ^{2} - 3 x - 4}

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

f (x) = x 25 - x^{2}, - 1 \leq x \leq 5

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