What is the extreme f(x)=(x^2+2)/(x^2-25) ?

The extreme f(x)=(x^2+2)/(x^2-25) is Maximum(0,-2/25)

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extreme f(x)=(x^2+2)/(x^2-25)

extreme points

f (x) = \frac{x ^{2} + 2}{x ^{2} - 25}

Maximum (0, - \frac{2}{25})

Solution steps

f^{'} (x) = - \frac{54 x}{( x ^{2} - 25 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = 0

into

\frac{x ^{2} + 2}{x ^{2} - 25} : - \frac{2}{25}

Maximum (0, - \frac{2}{25})

f (x) = \frac{4 e ^{4 x}}{4 x - 5}

f (x) = x + 3 + 6

f (x) = 2 x^{2} - 8 x + 5

f (x) = x^{4} (x - 1)^{2}

\frac{4 x ^{2} - 3 x ^{4}}{2 x}

What is the extreme f(x)=(x^2+2)/(x^2-25) ?
The extreme f(x)=(x^2+2)/(x^2-25) is Maximum(0,-2/25)