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

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

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

extreme

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

Solution steps

f^{'} (x) = \frac{6 x}{( 4 - x ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

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

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