What is the extreme sin(3x) ?

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

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extreme sin(3x)

extreme points

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) = \frac{2 x + 5}{x - 1}

f (x) = \frac{x ^{2} + 5 x + 5}{x + 1}

y = (x - 3)^{2}

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

x^{2} + 3

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