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

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

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

extreme points

y = \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})

f (x) = 9 x - 8

4 - (x - 2)^{2}

f (x) = x^{3} - 7 x^{2} + 15 x + 9

f (x) = \frac{x - 4}{x ^{2} + 13 x + 36}

f (x) = - 4 x + 7 y = 3

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