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

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

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

extreme points

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

Maximum (0, \frac{1}{2})

Solution steps

f^{'} (x) = - \frac{12 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} - 2}{x ^{2} - 4} : \frac{1}{2}

Maximum (0, \frac{1}{2})

P (a, b) = 2 a + 2 b

f (x) = 5 (x - 3)^{2} - 6 y^{2} + 4 x y

y = x^{4} - 8 x^{2}

2 x^{3} + x y^{2} + 5 x^{2} + y^{2} + 5

f (x) = - x^{3} + 2 x^{2} + 19 x - 20

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