What is the extreme f(x)=(-2(x^2-1))/(x^2-4) ?

The extreme f(x)=(-2(x^2-1))/(x^2-4) is Minimum(0,-1/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(-2(x^2-1))/(x^2-4)

extreme points

f (x) = \frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4}

Minimum (0, - \frac{1}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

\frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4} : - \frac{1}{2}

Minimum (0, - \frac{1}{2})

e^{- x^{2}} (4 x^{2} - 2)

f (x, y) = \frac{e ^{2 - x y}}{x + y ^{2}}

f (x) = \frac{- 7}{x ^{2} - 9}

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

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 3 y - 10

What is the extreme f(x)=(-2(x^2-1))/(x^2-4) ?
The extreme f(x)=(-2(x^2-1))/(x^2-4) is Minimum(0,-1/2)