What is the extreme (x^3)/(x^2+1) ?

The extreme (x^3)/(x^2+1) is Saddle(0,0)

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extreme (x^3)/(x^2+1)

extreme points

\frac{x ^{3}}{x ^{2} + 1}

Saddle (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

\frac{x ^{3}}{x ^{2} + 1} : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug the extreme point

x = 0

into

\frac{x ^{3}}{x ^{2} + 1} \Rightarrow y = 0

Saddle (0, 0)

y = x^{2} + 4 x + 4

f (x) = x

f (x) = \frac{1 - 4 x}{2 x + 9}

f (x) = 4 + 8 x - 5 x^{2}

f (x) = x + 9 - 2