What is the extreme y=sin(3x) ?

The extreme y=sin(3x) is Maximum(pi/6+(2pi)/3 n,1),Minimum(pi/2+(2pi)/3 n,-1)

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extreme y=sin(3x)

extreme points

y = sin (3 x)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 1), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 1)

Solution steps

Find the critical points:

x = \frac{π}{6} + \frac{2 π}{3} n, x = \frac{π}{2} + \frac{2 π}{3} n

Domain of

sin (3 x) : - \infty < x < \infty

Find intervals:

Increasing

: \frac{2 π}{3} n \leq x < \frac{π}{6} + \frac{2 π}{3} n,

Decreasing

: \frac{π}{6} + \frac{2 π}{3} n < x < \frac{π}{2} + \frac{2 π}{3} n,

Increasing

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

Plug the extreme point

x = \frac{π}{6} + \frac{2 π}{3} n

into

sin (3 x) \Rightarrow y = 1

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 1)

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

sin (3 x) \Rightarrow y = - 1

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

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 1), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 1)

f (x) = e^{4 x^{2} + y^{2} + 3}

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

\frac{x}{x ^{2} - 64}

f (x) = 2 ln (x) + ln (y) - 4 x - y

f (x) = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 8

What is the extreme y=sin(3x) ?
The extreme y=sin(3x) is Maximum(pi/6+(2pi)/3 n,1),Minimum(pi/2+(2pi)/3 n,-1)