What is the extreme f(x)=sin(6x) ?

The extreme f(x)=sin(6x) is Maximum(pi/(12)+pi/3 n,1),Minimum(pi/4+pi/3 n,-1)

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extreme f(x)=sin(6x)

extreme points

f (x) = sin (6 x)

Maximum (\frac{π}{12} + \frac{π}{3} n, 1), Minimum (\frac{π}{4} + \frac{π}{3} n, - 1)

Solution steps

Find the critical points:

x = \frac{π}{12} + \frac{π}{3} n, x = \frac{π}{4} + \frac{π}{3} n

Domain of

sin (6 x) : - \infty < x < \infty

Find intervals:

Increasing

: \frac{π}{3} n \leq x < \frac{π}{12} + \frac{π}{3} n,

Decreasing

: \frac{π}{12} + \frac{π}{3} n < x < \frac{π}{4} + \frac{π}{3} n,

Increasing

: \frac{π}{4} + \frac{π}{3} n < x < \frac{π}{3} + \frac{π}{3} n

Plug the extreme point

x = \frac{π}{12} + \frac{π}{3} n

into

sin (6 x) \Rightarrow y = 1

Maximum (\frac{π}{12} + \frac{π}{3} n, 1)

Plug the extreme point

x = \frac{π}{4} + \frac{π}{3} n

into

sin (6 x) \Rightarrow y = - 1

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

Maximum (\frac{π}{12} + \frac{π}{3} n, 1), Minimum (\frac{π}{4} + \frac{π}{3} n, - 1)

f (x) = x^{3} - \frac{3}{2} x^{2} - 6 x

f (x) = x^{\frac{3}{5}}

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

f (x, y) = - 6 x^{2} + 5 x y - y^{2} + x + y

f (x) = x^{\frac{5}{3}} - 12 x^{\frac{2}{3}}

What is the extreme f(x)=sin(6x) ?
The extreme f(x)=sin(6x) is Maximum(pi/(12)+pi/3 n,1),Minimum(pi/4+pi/3 n,-1)