What is the extreme f(x)=3+sin(pi/3 x) ?

The extreme f(x)=3+sin(pi/3 x) is Maximum(3/2+6n,4),Minimum(9/2+6n,2)

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extreme f(x)=3+sin(pi/3 x)

extreme points

f (x) = 3 + sin (\frac{π}{3} x)

Maximum (\frac{3}{2} + 6 n, 4), Minimum (\frac{9}{2} + 6 n, 2)

Solution steps

Find the critical points:

x = \frac{3}{2} + 6 n, x = \frac{9}{2} + 6 n

Domain of

3 + sin (\frac{π}{3} x) : - \infty < x < \infty

Find intervals:

Increasing

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

Decreasing

: \frac{3}{2} + 6 n < x < \frac{9}{2} + 6 n,

Increasing

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

Plug the extreme point

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

into

3 + sin (\frac{π}{3} x) \Rightarrow y = 4

Maximum (\frac{3}{2} + 6 n, 4)

Plug the extreme point

x = \frac{9}{2} + 6 n

into

3 + sin (\frac{π}{3} x) \Rightarrow y = 2

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

Maximum (\frac{3}{2} + 6 n, 4), Minimum (\frac{9}{2} + 6 n, 2)

2 x - y = 9

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

f (x) = \frac{( 2 z - 6 )}{9 z + 1}

u (x) = x + 1

f (x) = 6 x + 6

What is the extreme f(x)=3+sin(pi/3 x) ?
The extreme f(x)=3+sin(pi/3 x) is Maximum(3/2+6n,4),Minimum(9/2+6n,2)