What is the extreme f(x)=2(csc(x)+sec(x)) ?

The extreme f(x)=2(csc(x)+sec(x)) is Minimum(pi/4+2pin,4sqrt(2)),Maximum((5pi)/4+2pin,-4sqrt(2))

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extreme f(x)=2(csc(x)+sec(x))

extreme points

f (x) = 2 (csc (x) + sec (x))

Minimum (\frac{π}{4} + 2 πn, 42), Maximum (\frac{5 π}{4} + 2 πn, - 42)

Solution steps

Find the critical points:

x = \frac{π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Domain of

2 (csc (x) + sec (x)) : 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < π + 2 πn or π + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Find intervals:

Decreasing

: 2 πn < x < \frac{π}{4} + 2 πn,

Increasing

: \frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn,

Increasing

: \frac{π}{2} + 2 πn < x < π + 2 πn,

Increasing

: π + 2 πn < x < \frac{5 π}{4} + 2 πn,

Decreasing

: \frac{5 π}{4} + 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}{4} + 2 πn

into

2 (csc (x) + sec (x)) \Rightarrow y = 42

Minimum (\frac{π}{4} + 2 πn, 42)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

2 (csc (x) + sec (x)) \Rightarrow y = - 42

Maximum (\frac{5 π}{4} + 2 πn, - 42)

Minimum (\frac{π}{4} + 2 πn, 42), Maximum (\frac{5 π}{4} + 2 πn, - 42)

f (x, y) = x e^{y} - ln (x)

y = \frac{5}{x}

f (x) = x^{4} - x^{2} + 2

f (x) = \frac{x ^{2} - 16}{x - 5}

15 sin (5000 π t)

What is the extreme f(x)=2(csc(x)+sec(x)) ?
The extreme f(x)=2(csc(x)+sec(x)) is Minimum(pi/4+2pin,4sqrt(2)),Maximum((5pi)/4+2pin,-4sqrt(2))