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

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

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

extreme points

f (x) = \frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4}

Minimum (0, - \frac{1}{2})

Solution steps

f^{'} (x) = \frac{12 x}{( x ^{2} - 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, - \frac{1}{2})

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