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

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

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extreme f(x)=2sin(x)

extreme points

f (x) = 2 sin (x)

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

2 sin (x) \Rightarrow y = 2

Maximum (\frac{π}{2} + 2 πn, 2)

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

2 sin (x) \Rightarrow y = - 2

Minimum (\frac{3 π}{2} + 2 πn, - 2)

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

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

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

sin (x) - x

f (x, y) = y + 9 - x^{2} - y^{2}

f (x) = - x^{2} + 160 x - 400

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