What is the extreme f(x)=(6x-10)/(x^2-1) ?

The extreme f(x)=(6x-10)/(x^2-1) is Minimum(1/3 ,9),Maximum(3,1)

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extreme f(x)=(6x-10)/(x^2-1)

extreme points

f (x) = \frac{6 x - 10}{x ^{2} - 1}

Minimum (\frac{1}{3}, 9), Maximum (3, 1)

Solution steps

f^{'} (x) = \frac{- 6 x ^{2} + 20 x - 6}{( x ^{2} - 1 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < 1,

Increasing

: 1 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = \frac{1}{3}

into

\frac{6 x - 10}{x ^{2} - 1} : 9

Minimum (\frac{1}{3}, 9)

Plug

x = 3

into

\frac{6 x - 10}{x ^{2} - 1} : 1

Maximum (3, 1)

Minimum (\frac{1}{3}, 9), Maximum (3, 1)

f (x) = \frac{3}{7} x - 6

\frac{1}{- x + 4}

f (x) = x^{3} - 4 x^{2} - 16 x - 3

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

f (x) = - x^{2} - 1

What is the extreme f(x)=(6x-10)/(x^2-1) ?
The extreme f(x)=(6x-10)/(x^2-1) is Minimum(1/3 ,9),Maximum(3,1)