What is the extreme ((x^2+2))/((x^2-16)) ?

The extreme ((x^2+2))/((x^2-16)) is Maximum(0,-1/8)

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extreme ((x^2+2))/((x^2-16))

extreme points

\frac{( x ^{2} + 2 )}{( x ^{2} - 16 )}

Maximum (0, - \frac{1}{8})

Solution steps

f^{'} (x) = - \frac{36 x}{( x ^{2} - 16 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

\frac{x ^{2} + 2}{x ^{2} - 16} : - \frac{1}{8}

Maximum (0, - \frac{1}{8})

y = 2 x^{2} + 5

f (x) = 5 x^{4} - 20 x^{3} + 1

f (x) = 0.001 x^{2} + 3.2 x - 10

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

f (x) = - 7 x^{2} - 7 x y - 6 y^{2} - 35 x - 26 y + 9

What is the extreme ((x^2+2))/((x^2-16)) ?
The extreme ((x^2+2))/((x^2-16)) is Maximum(0,-1/8)