What is the extreme f(x)=(x-4)/(3x-x^2) ?

The extreme f(x)=(x-4)/(3x-x^2) is Maximum(2,-1),Minimum(6,-1/9)

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extreme f(x)=(x-4)/(3x-x^2)

extreme points

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

Maximum (2, - 1), Minimum (6, - \frac{1}{9})

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 12}{( 3 x - x ^{2} ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 3,

Decreasing

: 3 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

\frac{x - 4}{3 x - x ^{2}} : - 1

Maximum (2, - 1)

Plug

x = 6

into

\frac{x - 4}{3 x - x ^{2}} : - \frac{1}{9}

Minimum (6, - \frac{1}{9})

Maximum (2, - 1), Minimum (6, - \frac{1}{9})

What is the extreme f(x)=(x-4)/(3x-x^2) ?
The extreme f(x)=(x-4)/(3x-x^2) is Maximum(2,-1),Minimum(6,-1/9)