What is the extreme f(x)=|(x^2-7x-6)/(x-5)| ?

The extreme f(x)=|(x^2-7x-6)/(x-5)| is Minimum((7-sqrt(73))/2 ,0),Minimum((7+sqrt(73))/2 ,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=|(x^2-7x-6)/(x-5)|

extreme

Minimum (\frac{7 - 73}{2}, 0), Minimum (\frac{7 + 73}{2}, 0)

Solution steps

Find the critical points:

x = \frac{7 - 73}{2}, x = \frac{7 + 73}{2}

Find intervals:

Decreasing

: - \infty < x < \frac{7 - 73}{2},

Increasing

: \frac{7 - 73}{2} < x < 5,

Decreasing

: 5 < x < \frac{7 + 73}{2},

Increasing

: \frac{7 + 73}{2} < x < \infty

Plug the extreme point

x = \frac{7 - 73}{2}

into

\frac{x ^{2} - 7 x - 6}{x - 5} \Rightarrow y = 0

Minimum (\frac{7 - 73}{2}, 0)

Plug the extreme point

x = \frac{7 + 73}{2}

into

\frac{x ^{2} - 7 x - 6}{x - 5} \Rightarrow y = 0

Minimum (\frac{7 + 73}{2}, 0)

Minimum (\frac{7 - 73}{2}, 0), Minimum (\frac{7 + 73}{2}, 0)

What is the extreme f(x)=|(x^2-7x-6)/(x-5)| ?
The extreme f(x)=|(x^2-7x-6)/(x-5)| is Minimum((7-sqrt(73))/2 ,0),Minimum((7+sqrt(73))/2 ,0)