What is the extreme f(x)=cos(x)-9x,0<= x<= 4pi ?

The extreme f(x)=cos(x)-9x,0<= x<= 4pi is Maximum(0,1),Minimum(4pi,1-36pi)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=cos(x)-9x,0<= x<= 4pi

extreme

Maximum (0, 1), Minimum (4 π, 1 - 36 π)

Solution steps

Find the critical points:

x = 0, x = 4 π

Domain of

cos (x) - 9 x, 0 \leq x \leq 4 π : 0 \leq x \leq 4 π

Find intervals:

Decreasing

: 0 < x < 4 π

Plug the extreme point

x = 0

into

cos (x) - 9 x \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 4 π

into

cos (x) - 9 x \Rightarrow y = 1 - 36 π

Minimum (4 π, 1 - 36 π)

Maximum (0, 1), Minimum (4 π, 1 - 36 π)

What is the extreme f(x)=cos(x)-9x,0<= x<= 4pi ?
The extreme f(x)=cos(x)-9x,0<= x<= 4pi is Maximum(0,1),Minimum(4pi,1-36pi)