What is the extreme f(x)=((x^2-5))/(x+3) ?

The extreme f(x)=((x^2-5))/(x+3) is Maximum(-5,-10),Minimum(-1,-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=((x^2-5))/(x+3)

extreme

Maximum (- 5, - 10), Minimum (- 1, - 2)

Solution steps

f^{'} (x) = \frac{x ^{2} + 6 x + 5}{( x + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 5

into

\frac{x ^{2} - 5}{x + 3} : - 10

Maximum (- 5, - 10)

Plug

x = - 1

into

\frac{x ^{2} - 5}{x + 3} : - 2

Minimum (- 1, - 2)

Maximum (- 5, - 10), Minimum (- 1, - 2)

What is the extreme f(x)=((x^2-5))/(x+3) ?
The extreme f(x)=((x^2-5))/(x+3) is Maximum(-5,-10),Minimum(-1,-2)