What is the extreme f(x)=((x^3))/((x-1)^2) ?

The extreme f(x)=((x^3))/((x-1)^2) is Saddle(0,0),Minimum(3, 27/4)

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extreme f(x)=((x^3))/((x-1)^2)

extreme points

f (x) = \frac{( x ^{3} )}{( x - 1 ) ^{2}}

Saddle (0, 0), Minimum (3, \frac{27}{4})

Solution steps

f^{'} (x) = \frac{x ^{2} ( x - 3 )}{( x - 1 ) ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

\frac{x ^{3}}{( x - 1 ) ^{2}} : 0

Saddle (0, 0)

Plug

x = 3

into

\frac{x ^{3}}{( x - 1 ) ^{2}} : \frac{27}{4}

Minimum (3, \frac{27}{4})

Saddle (0, 0), Minimum (3, \frac{27}{4})

y = \frac{2 x ^{2}}{x - 6} [- 3]

f (x, y) = x - 2 y

3 x + 1 [3.6]

f (x) = \frac{9 x}{x - 4}, 6 \leq x \leq 12

What is the extreme f(x)=((x^3))/((x-1)^2) ?
The extreme f(x)=((x^3))/((x-1)^2) is Saddle(0,0),Minimum(3, 27/4)