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

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

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

extreme points

\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})

f (x, y) = x^{3} - y^{3} - 2 x y + 10

y = x^{4} - 18 x^{2}

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x + 1

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, (0, 10)

x^{2} y + x y^{2} + 3 x y

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