What is the extreme f(x)=2cos^2(x),(0,pi) ?

The extreme f(x)=2cos^2(x),(0,pi) is Minimum(pi/2 ,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=2cos^2(x),(0,pi)

extreme points

f (x) = 2 cos^{2} (x), (0, π)

Minimum (\frac{π}{2}, 0)

Solution steps

Find the critical points:

x = \frac{π}{2}

Domain of

2 cos^{2} (x), (0, π) : 0 < x < π

Find intervals:

Decreasing

: 0 < x < \frac{π}{2},

Increasing

: \frac{π}{2} < x < π

Plug the extreme point

x = \frac{π}{2}

into

2 cos^{2} (x) \Rightarrow y = 0

Minimum (\frac{π}{2}, 0)

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

f (x) = 1.5 x^{2} + 45 x + 15000

f (x) = x^{3} - 8 x^{2} + 3

f (x) = x^{3} - 12 x + 18

f (x) = 2 x^{3} - 20 x + 50 x - 6

What is the extreme f(x)=2cos^2(x),(0,pi) ?
The extreme f(x)=2cos^2(x),(0,pi) is Minimum(pi/2 ,0)