What is the extreme f(x)=4csc(x) ?

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

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extreme f(x)=4csc(x)

extreme points

f (x) = 4 csc (x)

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

Solution steps

Find the critical points:

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Domain of

4 csc (x) : 2 πn < x < π + 2 πn or π + 2 πn < x < 2 π + 2 πn

Find intervals:

Decreasing

: 2 πn < x < \frac{π}{2} + 2 πn,

Increasing

: \frac{π}{2} + 2 πn < x < π + 2 πn,

Increasing

: π + 2 πn < x < \frac{3 π}{2} + 2 πn,

Decreasing

: \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

4 csc (x) \Rightarrow y = 4

Minimum (\frac{π}{2} + 2 πn, 4)

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

4 csc (x) \Rightarrow y = - 4

Maximum (\frac{3 π}{2} + 2 πn, - 4)

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

S (v, t) = v \cdot t - \frac{v}{t} \cdot \frac{2 t ^{2}}{2}

f (x) = 2 x^{3} - 96 x - 3

x ln (\frac{x}{9}) - x

f (x) = x^{3} + x^{2} - 6 x

f (x, y) = x^{2} - 2 x + y^{2} + 1

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