What is the extreme f(x)=2sin(3x-18)+3 ?

The extreme f(x)=2sin(3x-18)+3 is Maximum((36-11pi)/6+(2pi)/3 n,5),Minimum((12-3pi)/2+(2pi)/3 n,1)

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extreme f(x)=2sin(3x-18)+3

extreme points

f (x) = 2 sin (3 x - 18) + 3

Maximum (\frac{36 - 11 π}{6} + \frac{2 π}{3} n, 5), Minimum (\frac{12 - 3 π}{2} + \frac{2 π}{3} n, 1)

Solution steps

Find the critical points:

x = \frac{36 - 11 π}{6} + \frac{2 π}{3} n, x = \frac{12 - 3 π}{2} + \frac{2 π}{3} n

Domain of

2 sin (3 x - 18) + 3 : - \infty < x < \infty

Find intervals:

Increasing

: \frac{2 π}{3} n \leq x < \frac{36 - 11 π}{6} + \frac{2 π}{3} n,

Decreasing

: \frac{36 - 11 π}{6} + \frac{2 π}{3} n < x < \frac{12 - 3 π}{2} + \frac{2 π}{3} n,

Increasing

: \frac{12 - 3 π}{2} + \frac{2 π}{3} n < x < \frac{2 π}{3} + \frac{2 π}{3} n

Plug the extreme point

x = \frac{36 - 11 π}{6} + \frac{2 π}{3} n

into

2 sin (3 x - 18) + 3 \Rightarrow y = 5

Maximum (\frac{36 - 11 π}{6} + \frac{2 π}{3} n, 5)

Plug the extreme point

x = \frac{12 - 3 π}{2} + \frac{2 π}{3} n

into

2 sin (3 x - 18) + 3 \Rightarrow y = 1

Minimum (\frac{12 - 3 π}{2} + \frac{2 π}{3} n, 1)

Maximum (\frac{36 - 11 π}{6} + \frac{2 π}{3} n, 5), Minimum (\frac{12 - 3 π}{2} + \frac{2 π}{3} n, 1)

What is the extreme f(x)=2sin(3x-18)+3 ?
The extreme f(x)=2sin(3x-18)+3 is Maximum((36-11pi)/6+(2pi)/3 n,5),Minimum((12-3pi)/2+(2pi)/3 n,1)