What is the extreme f(x)=(12x-3-3x^2)/x ?

The extreme f(x)=(12x-3-3x^2)/x is Minimum(-1,18),Maximum(1,6)

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extreme f(x)=(12x-3-3x^2)/x

extreme points

f (x) = \frac{12 x - 3 - 3 x ^{2}}{x}

Minimum (- 1, 18), Maximum (1, 6)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{12 x - 3 - 3 x ^{2}}{x} : 18

Minimum (- 1, 18)

Plug

x = 1

into

\frac{12 x - 3 - 3 x ^{2}}{x} : 6

Maximum (1, 6)

Minimum (- 1, 18), Maximum (1, 6)

f (x) = - x^{2} + 12

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

f (x, y) = x^{2} + y^{2} - 12 x + 16 y

f (x) = 14 x + 15 x^{\frac{14}{15}}

f (x) = \frac{x ^{3}}{3} + \frac{15 x ^{2}}{2} + 50 x - 2

What is the extreme f(x)=(12x-3-3x^2)/x ?
The extreme f(x)=(12x-3-3x^2)/x is Minimum(-1,18),Maximum(1,6)