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)

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

extreme points

f (x) = \frac{x ^{2} - 7 x - 6}{x - 5}

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)

(x + 1)^{3} (x - 3)

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 6, - 5 \leq x \leq 5

f (x) = x^{\frac{1}{7}}

f (x, y) = x^{2} + 2 x - y^{2}

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)