What is the extreme f(x)=(2(-2x^2+4))/(sqrt(4-x^2)) ?

The extreme f(x)=(2(-2x^2+4))/(sqrt(4-x^2)) is Maximum(0,4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(2(-2x^2+4))/(sqrt(4-x^2))

extreme points

f (x) = \frac{2 ( - 2 x ^{2} + 4 )}{4 - x ^{2}}

Maximum (0, 4)

Solution steps

f^{'} (x) = \frac{2 ( 2 x ^{3} - 12 x )}{( 4 - x ^{2} ) 4 - x ^{2}}

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2

Plug

x = 0

into

\frac{2 ( - 2 x ^{2} + 4 )}{4 - x ^{2}} : 4

Maximum (0, 4)

x^{2} + x

5 x y + 4 ln (x) + 10 y

f (x, y) = x^{2} + y^{2} + 2 x - 4 y + 2

f (x) = 3 x^{4} + 8 x^{3} - 14 x^{2} - 24 x + 17

f (x, y) = \frac{x + y}{x + y}

What is the extreme f(x)=(2(-2x^2+4))/(sqrt(4-x^2)) ?
The extreme f(x)=(2(-2x^2+4))/(sqrt(4-x^2)) is Maximum(0,4)