What is the extreme f(x)=sqrt(x^2-9) ?

The extreme f(x)=sqrt(x^2-9) is Minimum(-3,0),Minimum(3,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=sqrt(x^2-9)

extreme points

f (x) = x^{2} - 9

Minimum (- 3, 0), Minimum (3, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{2} - 9 : 0

Minimum (- 3, 0)

Plug

x = 3

into

x^{2} - 9 : 0

Minimum (3, 0)

Minimum (- 3, 0), Minimum (3, 0)

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

f (x) = x^{2} - 4

\frac{d}{d x} (I n \frac{x + 1}{x - 1})

x^{4} - 2 x^{3} + 2

f (x) = \frac{x ^{3}}{x ^{2} - 9}

What is the extreme f(x)=sqrt(x^2-9) ?
The extreme f(x)=sqrt(x^2-9) is Minimum(-3,0),Minimum(3,0)