What is the extreme f(x)=(2x^2+1)/(x^2-4) ?

The extreme f(x)=(2x^2+1)/(x^2-4) is Maximum(0,-1/4)

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extreme f(x)=(2x^2+1)/(x^2-4)

extreme

Maximum (0, - \frac{1}{4})

Solution steps

f^{'} (x) = - \frac{18 x}{( x ^{2} - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2} + 1}{x ^{2} - 4} : - \frac{1}{4}

Maximum (0, - \frac{1}{4})

What is the extreme f(x)=(2x^2+1)/(x^2-4) ?
The extreme f(x)=(2x^2+1)/(x^2-4) is Maximum(0,-1/4)