What is the extreme f(x)=(2x^2+5)/(x^2-25) ?

The extreme f(x)=(2x^2+5)/(x^2-25) is Maximum(0,-1/5)

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extreme f(x)=(2x^2+5)/(x^2-25)

extreme points

f (x) = \frac{2 x ^{2} + 5}{x ^{2} - 25}

Maximum (0, - \frac{1}{5})

Solution steps

f^{'} (x) = - \frac{110 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{2 x ^{2} + 5}{x ^{2} - 25} : - \frac{1}{5}

Maximum (0, - \frac{1}{5})

f (x, y) = x - 5 y + 6

f (x) = 2 x - 6 x^{\frac{1}{3}}

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 11

f (x) = (1 + \frac{20}{x}) x^{2} + 144

4 sin (x) + 4 cos (x), 0 \leq x \leq \frac{π}{2}

What is the extreme f(x)=(2x^2+5)/(x^2-25) ?
The extreme f(x)=(2x^2+5)/(x^2-25) is Maximum(0,-1/5)