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

The extreme f(x)=sin(4x) is Maximum(pi/8+pi/2 n,1),Minimum((3pi)/8+pi/2 n,-1)

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extreme f(x)=sin(4x)

extreme points

f (x) = sin (4 x)

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

Solution steps

Find the critical points:

x = \frac{π}{8} + \frac{π}{2} n, x = \frac{3 π}{8} + \frac{π}{2} n

Domain of

sin (4 x) : - \infty < x < \infty

Find intervals:

Increasing

: \frac{π}{2} n \leq x < \frac{π}{8} + \frac{π}{2} n,

Decreasing

: \frac{π}{8} + \frac{π}{2} n < x < \frac{3 π}{8} + \frac{π}{2} n,

Increasing

: \frac{3 π}{8} + \frac{π}{2} n < x < \frac{π}{2} + \frac{π}{2} n

Plug the extreme point

x = \frac{π}{8} + \frac{π}{2} n

into

sin (4 x) \Rightarrow y = 1

Maximum (\frac{π}{8} + \frac{π}{2} n, 1)

Plug the extreme point

x = \frac{3 π}{8} + \frac{π}{2} n

into

sin (4 x) \Rightarrow y = - 1

Minimum (\frac{3 π}{8} + \frac{π}{2} n, - 1)

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

f (x) = x + 2 y

f (x) = - x^{3} + 3 x - 7

f (x, y) = x^{2} + 4 y^{2} - 6 x + 16 y

f (x, y) = 8 x y - x^{3} - 4 y^{2}

f (x) = 5 x^{6} - 3 x^{4} + 2 x - 9

What is the extreme f(x)=sin(4x) ?
The extreme f(x)=sin(4x) is Maximum(pi/8+pi/2 n,1),Minimum((3pi)/8+pi/2 n,-1)