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

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

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

extreme points

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

Saddle (0, \frac{1}{3})

Solution steps

Find the critical points:

x = 0

Domain of

Find intervals:

Decreasing

: 0 < x < \infty

Plug the extreme point

x = 0

into

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

Saddle (0, \frac{1}{3})

y = x - 2 cos (x), 0 \leq x \leq 2 π

f (x, y) = x^{2} - 10 y^{2}

6 x - 6 x

f (x, y) = 11 x^{3} - 33 x y + 11 y^{3}

f (x) = x^{3} + y^{2} - 3 x^{2} - 3 y^{2} - 9 x

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