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

The extreme f(x)=|(x-2)/(x+5)|-3 is Minimum(2,-3)

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)/(x+5)|-3

extreme points

f (x) = \frac{x - 2}{x + 5} - 3

Minimum (2, - 3)

Solution steps

Find the critical points:

x = 2

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 2

into

\frac{x - 2}{x + 5} - 3 \Rightarrow y = - 3

Minimum (2, - 3)

2 x^{4} - x^{2} + 5

f (x) = 7 x^{9} - 9 x^{7} - 5 (- 3.4)

f (x) = - 3 + x^{2} - 3 x - 3 x^{3} + 2 x^{5} + 4 x^{4}

f (x) = - x^{3} + 6 x^{2} + 135 x + 1

x^{2} - 8 x + 15

What is the extreme f(x)=|(x-2)/(x+5)|-3 ?
The extreme f(x)=|(x-2)/(x+5)|-3 is Minimum(2,-3)