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

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

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

extreme

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})

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