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

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

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

extreme points

f (x) = \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) = \frac{1}{3} x^{3} - 9 x + 2

f (x) = e^{x^{2} - 7 x - 1}

f (x) = 25 - x^{2} - y^{2}

f (x) = \frac{x ^{2} + 2 x + 1}{x}

f (x) = x^{2} - 1

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