What is the extreme f(x)=3cos(x-pi/2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3cos(x-pi/2)

extreme points

f (x) = 3 cos (x - \frac{π}{2})

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

Solution steps

Find the critical points:

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

Domain of

3 cos (x - \frac{π}{2}) : - \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

3 cos (x - \frac{π}{2}) \Rightarrow y = 3

Maximum (\frac{π}{2} + 2 πn, 3)

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

3 cos (x - \frac{π}{2}) \Rightarrow y = - 3

Minimum (\frac{3 π}{2} + 2 πn, - 3)

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

Q (a, b) = 7 a b^{2} - 6 a^{3} b + 12

f (x) = - 2 x^{2} + 8 x

f (x) = 3 - 3 x^{2}, - 5 \leq x \leq 2

f (x) = \frac{2}{5 x ^{2} + 1}, - 2 \leq x \leq 5

f (x) = - 1.872 x^{2} + 33.792 - 102.353

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