What is the extreme f(x)=4sin(2x) ?

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

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extreme f(x)=4sin(2x)

extreme points

f (x) = 4 sin (2 x)

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

Solution steps

Find the critical points:

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

Domain of

4 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

4 sin (2 x) \Rightarrow y = 4

Maximum (\frac{π}{4} + πn, 4)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

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

Minimum (\frac{3 π}{4} + πn, - 4)

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

f (x) = \frac{( 2 x ^{2} + x - 3 )}{( 3 x - 9 )}

f (x, y) = x^{2} - y^{2} - 20 x + 2 y + 8

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

x^{4} + 2 x^{3} + 1

f (x, y) = x^{2} + 3 y^{2} - 2 x y + 10 x - 2 y + 4

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