What is the extreme f(x)=(x^2+4)/(x^2-25) ?

The extreme f(x)=(x^2+4)/(x^2-25) is Maximum(0,-4/25)

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extreme f(x)=(x^2+4)/(x^2-25)

extreme points

f (x) = \frac{x ^{2} + 4}{x ^{2} - 25}

Maximum (0, - \frac{4}{25})

Solution steps

f^{'} (x) = - \frac{58 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} + 4}{x ^{2} - 25} : - \frac{4}{25}

Maximum (0, - \frac{4}{25})

f (x) = (x - 8) ln (x - 8)

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

f (x) = x^{3} + 6 x^{2} + 12 x + 5

f (x) = x^{4} - 98 x^{2} + 4

(x + 3)^{2} (x^{3} - x^{2})

What is the extreme f(x)=(x^2+4)/(x^2-25) ?
The extreme f(x)=(x^2+4)/(x^2-25) is Maximum(0,-4/25)