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

The extreme f(x)=(x^3)/(x-3) is Saddle(0,0),Minimum(9/2 , 243/4)

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

extreme points

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

Saddle (0, 0), Minimum (\frac{9}{2}, \frac{243}{4})

Solution steps

f^{'} (x) = \frac{2 x ^{3} - 9 x ^{2}}{( x - 3 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Decreasing

: 3 < x < \frac{9}{2},

Increasing

: \frac{9}{2} < x < \infty

Plug

x = 0

into

\frac{x ^{3}}{x - 3} : 0

Saddle (0, 0)

Plug

x = \frac{9}{2}

into

\frac{x ^{3}}{x - 3} : \frac{243}{4}

Minimum (\frac{9}{2}, \frac{243}{4})

Saddle (0, 0), Minimum (\frac{9}{2}, \frac{243}{4})

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

5 x - 2

2 x + 3

g (x) = x + 3

f (x) = 3 x^{2} - 27

What is the extreme f(x)=(x^3)/(x-3) ?
The extreme f(x)=(x^3)/(x-3) is Saddle(0,0),Minimum(9/2 , 243/4)