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

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

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extreme y=((x^2+1))/((x^2-4))

extreme points

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

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

Solution steps

f^{'} (x) = - \frac{10 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{x ^{2} + 1}{x ^{2} - 4} : - \frac{1}{4}

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

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

f (x, y) = - x^{2} - y^{2} - 2 y + 5

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 4

f (x) = x^{2} y + y^{3} - 12 y

\frac{e ^{x}}{x + 1}

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