What is the extreme f(x)=5-3sin(2x) ?

The extreme f(x)=5-3sin(2x) is Minimum(pi/4+pin,2),Maximum((3pi)/4+pin,8)

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extreme f(x)=5-3sin(2x)

extreme points

f (x) = 5 - 3 sin (2 x)

Minimum (\frac{π}{4} + πn, 2), Maximum (\frac{3 π}{4} + πn, 8)

Solution steps

Find the critical points:

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Domain of

5 - 3 sin (2 x) : - \infty < x < \infty

Find intervals:

Decreasing

: πn \leq x < \frac{π}{4} + πn,

Increasing

: \frac{π}{4} + πn < x < \frac{3 π}{4} + πn,

Decreasing

: \frac{3 π}{4} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{4} + πn

into

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

Minimum (\frac{π}{4} + πn, 2)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

5 - 3 sin (2 x) \Rightarrow y = 8

Maximum (\frac{3 π}{4} + πn, 8)

Minimum (\frac{π}{4} + πn, 2), Maximum (\frac{3 π}{4} + πn, 8)

\frac{2 x - 4}{3}

f (x, y) = 8 y^{2} + x^{2} - x^{2} y

f (x) = x^{3} - 3 x, 0 \leq x \leq 2

2 x - x

f (x) = 2 sin (x) - x

What is the extreme f(x)=5-3sin(2x) ?
The extreme f(x)=5-3sin(2x) is Minimum(pi/4+pin,2),Maximum((3pi)/4+pin,8)