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

The extreme f(x)=-sin(x) is Minimum(pi/2+2pin,-1),Maximum((3pi)/2+2pin,1)

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extreme f(x)=-sin(x)

extreme points

f (x) = - sin (x)

Minimum (\frac{π}{2} + 2 πn, - 1), Maximum (\frac{3 π}{2} + 2 πn, 1)

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Increasing

: \frac{π}{2} + 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

- sin (x) \Rightarrow y = - 1

Minimum (\frac{π}{2} + 2 πn, - 1)

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

- sin (x) \Rightarrow y = 1

Maximum (\frac{3 π}{2} + 2 πn, 1)

Minimum (\frac{π}{2} + 2 πn, - 1), Maximum (\frac{3 π}{2} + 2 πn, 1)

f (x) = \frac{x}{ln ( x )} a t [2.5]

f (x) = (x + 5)^{3}

f (x) = x^{6} (1 - x)^{3}

f (x) = 6 x^{2} - 6 x - 12

f (x) = x^{2} + x y + y^{2} - 25 y + 208

What is the extreme f(x)=-sin(x) ?
The extreme f(x)=-sin(x) is Minimum(pi/2+2pin,-1),Maximum((3pi)/2+2pin,1)