What is the extreme g(x)=(x^3)/((x+1)) ?

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

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extreme g(x)=(x^3)/((x+1))

extreme points

g (x) = \frac{x ^{3}}{( x + 1 )}

Minimum (- \frac{3}{2}, \frac{27}{4}), Saddle (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{3}{2}

into

\frac{x ^{3}}{x + 1} : \frac{27}{4}

Minimum (- \frac{3}{2}, \frac{27}{4})

Plug

x = 0

into

\frac{x ^{3}}{x + 1} : 0

Saddle (0, 0)

Minimum (- \frac{3}{2}, \frac{27}{4}), Saddle (0, 0)

f (x) = x^{\frac{5}{4}} - 80 x^{\frac{1}{4}}

f (x) = 5 + 6 x - x^{2}, 0 \leq x \leq 4

f (x) = x^{3} + 3 x^{2} + y^{2} - 2 y + 3

f (x) = - x^{3} + 8 x^{2} - 15 x

y = 2 x^{3} - 15 x^{2} + 24 x - 5

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