What is the extreme f(x)=(4x^2+12)/(x-1) ?

The extreme f(x)=(4x^2+12)/(x-1) is Maximum(-1,-8),Minimum(3,24)

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extreme f(x)=(4x^2+12)/(x-1)

extreme points

f (x) = \frac{4 x ^{2} + 12}{x - 1}

Maximum (- 1, - 8), Minimum (3, 24)

Solution steps

f^{'} (x) = \frac{4 x ^{2} - 8 x - 12}{( 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{4 x ^{2} + 12}{x - 1} : - 8

Maximum (- 1, - 8)

Plug

x = 3

into

\frac{4 x ^{2} + 12}{x - 1} : 24

Minimum (3, 24)

Maximum (- 1, - 8), Minimum (3, 24)

f (x, y) = 4 y^{2} e^{y} + 6 x^{2}

f (x) = x^{4} - 2 x^{2} + x - 2

f (x) = x^{2} - 2 x - 9, 0 \leq x \leq 3

f (x) = tan (\frac{π x}{32}), 0 \leq x \leq 8

3 x^{3} + 9 x^{2} + 27 x - 21

What is the extreme f(x)=(4x^2+12)/(x-1) ?
The extreme f(x)=(4x^2+12)/(x-1) is Maximum(-1,-8),Minimum(3,24)