What is the extreme f(x)=((x^2))/(x^2+243) ?

The extreme f(x)=((x^2))/(x^2+243) is Minimum(0,0)

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extreme f(x)=((x^2))/(x^2+243)

extreme

Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{486 x}{( x ^{2} + 243 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x ^{2} + 243} : 0

Minimum (0, 0)

What is the extreme f(x)=((x^2))/(x^2+243) ?
The extreme f(x)=((x^2))/(x^2+243) is Minimum(0,0)