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

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

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

extreme points

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

Maximum (0, 2)

Solution steps

f^{'} (x) = - \frac{8 x}{( x + 2 ) ^{2} ( x - 2 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 2)

f (x, y) = e^{2 - x^{2} - y^{2}}

f (x) = - 12 x^{2} + 100

f (x) = 2 (x - 1) (x - 2) (x + 3)

C (x) = 30 x + 3300 I (x) - 10 x^{2} + 400 x

f (x, y) = x^{2} + 16 y^{2}

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