What is the extreme f(x)=(x^2+1)/(x^2-25) ?

The extreme f(x)=(x^2+1)/(x^2-25) is Maximum(0,-1/25)

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extreme f(x)=(x^2+1)/(x^2-25)

extreme points

f (x) = \frac{x ^{2} + 1}{x ^{2} - 25}

Maximum (0, - \frac{1}{25})

Solution steps

f^{'} (x) = - \frac{52 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} + 1}{x ^{2} - 25} : - \frac{1}{25}

Maximum (0, - \frac{1}{25})

2 x^{3} - x^{2} - 4 x + 4

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

2 x^{3} - x^{2} - 4 x + 8

f (x) = x^{3} - 13 x^{2} - 9 x

3 x^{3} - 9 x + 9 x y^{2}

What is the extreme f(x)=(x^2+1)/(x^2-25) ?
The extreme f(x)=(x^2+1)/(x^2-25) is Maximum(0,-1/25)