What is the extreme f(x)=sin^2(x/4) ?

The extreme f(x)=sin^2(x/4) is Minimum(4pin,0),Maximum(2pi+4pin,1)

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extreme f(x)=sin^2(x/4)

extreme points

f (x) = sin^{2} (\frac{x}{4})

Minimum (4 πn, 0), Maximum (2 π + 4 πn, 1)

Solution steps

Find the critical points:

x = 4 πn, x = 2 π + 4 πn

Domain of

sin^{2} (\frac{x}{4}) : - \infty < x < \infty

Find intervals:

Increasing

: 4 πn < x < 2 π + 4 πn,

Decreasing

: 2 π + 4 πn < x < 4 π + 4 πn

Plug the extreme point

x = 4 πn

into

sin^{2} (\frac{x}{4}) \Rightarrow y = 0

Minimum (4 πn, 0)

Plug the extreme point

x = 2 π + 4 πn

into

sin^{2} (\frac{x}{4}) \Rightarrow y = 1

Maximum (2 π + 4 πn, 1)

Minimum (4 πn, 0), Maximum (2 π + 4 πn, 1)

x - 3 y = - 3

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

\frac{1}{3} lo g_{10} (3 x)

f (x) = \frac{x - 2}{x ^{2} - 4}

What is the extreme f(x)=sin^2(x/4) ?
The extreme f(x)=sin^2(x/4) is Minimum(4pin,0),Maximum(2pi+4pin,1)