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

The extreme f(x)=|sin(x)| is Maximum(pi/2+2pin,1),Minimum(pi+2pin,0),Maximum((3pi)/2+2pin,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=|sin(x)|

extreme

Maximum (\frac{π}{2} + 2 πn, 1), Minimum (π + 2 πn, 0), Maximum (\frac{3 π}{2} + 2 πn, 1)

Solution steps

Find the critical points:

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

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Decreasing

: \frac{π}{2} + 2 πn < x < π + 2 πn,

Increasing

: π + 2 πn < x < \frac{3 π}{2} + 2 πn

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

∣ sin (x) ∣ \Rightarrow y = 1

Maximum (\frac{π}{2} + 2 πn, 1)

Plug the extreme point

x = π + 2 πn

into

∣ sin (x) ∣ \Rightarrow y = 0

Minimum (π + 2 πn, 0)

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)

Maximum (\frac{π}{2} + 2 πn, 1), Minimum (π + 2 πn, 0), Maximum (\frac{3 π}{2} + 2 πn, 1)

What is the extreme f(x)=|sin(x)| ?
The extreme f(x)=|sin(x)| is Maximum(pi/2+2pin,1),Minimum(pi+2pin,0),Maximum((3pi)/2+2pin,1)