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

The extreme f(x)=sin(9x) is Maximum(pi/(18)+(2pi)/9 n,1),Minimum(pi/6+(2pi)/9 n,-1)

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extreme f(x)=sin(9x)

extreme points

f (x) = sin (9 x)

Maximum (\frac{π}{18} + \frac{2 π}{9} n, 1), Minimum (\frac{π}{6} + \frac{2 π}{9} n, - 1)

Solution steps

Find the critical points:

x = \frac{π}{18} + \frac{2 π}{9} n, x = \frac{π}{6} + \frac{2 π}{9} n

Domain of

sin (9 x) : - \infty < x < \infty

Find intervals:

Increasing

: \frac{2 π}{9} n \leq x < \frac{π}{18} + \frac{2 π}{9} n,

Decreasing

: \frac{π}{18} + \frac{2 π}{9} n < x < \frac{π}{6} + \frac{2 π}{9} n,

Increasing

: \frac{π}{6} + \frac{2 π}{9} n < x < \frac{2 π}{9} + \frac{2 π}{9} n

Plug the extreme point

x = \frac{π}{18} + \frac{2 π}{9} n

into

sin (9 x) \Rightarrow y = 1

Maximum (\frac{π}{18} + \frac{2 π}{9} n, 1)

Plug the extreme point

x = \frac{π}{6} + \frac{2 π}{9} n

into

sin (9 x) \Rightarrow y = - 1

Minimum (\frac{π}{6} + \frac{2 π}{9} n, - 1)

Maximum (\frac{π}{18} + \frac{2 π}{9} n, 1), Minimum (\frac{π}{6} + \frac{2 π}{9} n, - 1)

s l o p e in t erce pt x + 2 y = 2

a sy m pt o t es \frac{( - x ^{2} + 7 x - 12 )}{5 x - 15}

in v erse f (x) = 6 - 9 x

in t erce pt s 6 sin (x)

d i s t an ce (4, - 3), (- 1, 1)

What is the extreme f(x)=sin(9x) ?
The extreme f(x)=sin(9x) is Maximum(pi/(18)+(2pi)/9 n,1),Minimum(pi/6+(2pi)/9 n,-1)