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

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

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extreme f(x)=sin(2x)

extreme points

f (x) = 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)

3 x^{2} + 10 x - 8

sin^{2} (x)

x 1 - x^{2}

f (x) = \frac{1}{x - 5}

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