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

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

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

extreme

Saddle (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

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

Saddle (0, 0)

What is the extreme f(x)=(x^3)/(3x^2+1) ?
The extreme f(x)=(x^3)/(3x^2+1) is Saddle(0,0)