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 points

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

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)

f (x) = sec^{2} (x) - 23 tan (x)

f (x) = 4 x^{3} - 12 x^{2} - 96 x + 9

f (x) = 4 x^{3} - 12 x^{2} - 96 x + 7

f (x, y) = - 25 x^{2} + 250 x - y^{2} + 8 y - 637

f (x, y) = 5 x^{2} y - 5 x y^{3} + 4 y^{3}

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