What is the extreme f(x)=(2x-2)/(x^2-10x+25) ?

The extreme f(x)=(2x-2)/(x^2-10x+25) is Minimum(-3,-1/8)

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extreme f(x)=(2x-2)/(x^2-10x+25)

extreme points

f (x) = \frac{2 x - 2}{x ^{2} - 10 x + 25}

Minimum (- 3, - \frac{1}{8})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = - 3

into

\frac{2 x - 2}{x ^{2} - 10 x + 25} : - \frac{1}{8}

Minimum (- 3, - \frac{1}{8})

f (x, y) = 5 x y

f (x) = \frac{2}{1 + x ^{2}}

\frac{x ^{2} - x - 6}{2 x + 4}

x x + 3

f (x, y) = 4 x y

What is the extreme f(x)=(2x-2)/(x^2-10x+25) ?
The extreme f(x)=(2x-2)/(x^2-10x+25) is Minimum(-3,-1/8)