What is the extreme f(x)=(x^2+1)/(x^2-25) ?

The extreme f(x)=(x^2+1)/(x^2-25) is Maximum(0,-1/25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(x^2+1)/(x^2-25)

extreme

Maximum (0, - \frac{1}{25})

Solution steps

f^{'} (x) = - \frac{52 x}{( x ^{2} - 25 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = 0

into

\frac{x ^{2} + 1}{x ^{2} - 25} : - \frac{1}{25}

Maximum (0, - \frac{1}{25})

What is the extreme f(x)=(x^2+1)/(x^2-25) ?
The extreme f(x)=(x^2+1)/(x^2-25) is Maximum(0,-1/25)