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)

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})

s l o p e in t erce pt x - 2 y = 2

v er t i ces y = - x^{2} + 2 x - 3

cr i t i c a l g (t) = \frac{4 t}{t ^{2} + 25}

in v erse f (x) = \frac{( 4 x + 5 )}{x + 4}

in v erse f (x) = {(- 5, 1), (4, 1)}

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)