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

The extreme f(x)=(2x^2+6)/(x-1) is Maximum(-1,-4),Minimum(3,12)

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

extreme points

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

Maximum (- 1, - 4), Minimum (3, 12)

Solution steps

f^{'} (x) = \frac{2 x ^{2} - 4 x - 6}{( x - 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

\frac{2 x ^{2} + 6}{x - 1} : - 4

Maximum (- 1, - 4)

Plug

x = 3

into

\frac{2 x ^{2} + 6}{x - 1} : 12

Minimum (3, 12)

Maximum (- 1, - 4), Minimum (3, 12)

f (x) = x^{5} + 32

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

P (x, y) = x y

x - 2 sin (x) (0.2 π)

f (x, y) = (x + y)^{- 1}

What is the extreme f(x)=(2x^2+6)/(x-1) ?
The extreme f(x)=(2x^2+6)/(x-1) is Maximum(-1,-4),Minimum(3,12)