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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

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

Saddle (0, 0)

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

f (x) = 4 x y - x^{3} - 2 y^{2}

f (x) = x^{2} - 8 ln (x), 1 \leq x \leq 5

\frac{x + 5}{x ^{2} - 2 x - 35}

f (x) = x^{2} 5 - x

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