What is the extreme sin(2x) ?

The extreme sin(2x) is Maximum(pi/4+pin,1),Minimum((3pi)/4+pin,-1)

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extreme sin(2x)

extreme points

sin (2 x)

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

Solution steps

Find the critical points:

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Domain of

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

Find intervals:

Increasing

: πn \leq x < \frac{π}{4} + πn,

Decreasing

: \frac{π}{4} + πn < x < \frac{3 π}{4} + πn,

Increasing

: \frac{3 π}{4} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{4} + πn

into

sin (2 x) \Rightarrow y = 1

Maximum (\frac{π}{4} + πn, 1)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

sin (2 x) \Rightarrow y = - 1

Minimum (\frac{3 π}{4} + πn, - 1)

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

f (x, y) = \frac{1}{3} x^{3} - 5 x^{2} - y^{2} - 3 y

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

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

g (x) = x (x - 1)^{2}

y = x^{3} - 12 x

What is the extreme sin(2x) ?
The extreme sin(2x) is Maximum(pi/4+pin,1),Minimum((3pi)/4+pin,-1)