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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Minimum (0, - 1)

Solution steps

f^{'} (x) = \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{2} - 4}{x ^{2} + 4} : - 1

Minimum (0, - 1)

f (x) = x \cdot ln (x)

f (x, y) = 2 x^{2} + 3 x y + y^{2} + 0 x + 5

f (x, y) = y^{2} (x - 1)^{2} + x^{2} (y + 4)^{2}

f (x) = (x - 2)^{2} + 1

2 x^{3} - 3 x^{2} - 12 x + 1

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