What is the extreme f(x)=9sin(3x) ?

The extreme f(x)=9sin(3x) is Maximum(pi/6+(2pi)/3 n,9),Minimum(pi/2+(2pi)/3 n,-9)

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extreme f(x)=9sin(3x)

extreme points

f (x) = 9 sin (3 x)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 9), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 9)

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

9 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

9 sin (3 x) \Rightarrow y = 9

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 9)

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

9 sin (3 x) \Rightarrow y = - 9

Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 9)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 9), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 9)

f (x, y) = 16 - 8 x + 10 y

f (x) = 20 sin (2 x)

x (3 - x)

f (x) = y \cdot e^{x} + x \cdot ln (y)

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

What is the extreme f(x)=9sin(3x) ?
The extreme f(x)=9sin(3x) is Maximum(pi/6+(2pi)/3 n,9),Minimum(pi/2+(2pi)/3 n,-9)