What is the extreme f(x)=-(x^2-50)/(x^2-25) ?

The extreme f(x)=-(x^2-50)/(x^2-25) is Maximum(0,-2)

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extreme f(x)=-(x^2-50)/(x^2-25)

extreme points

f (x) = - \frac{x ^{2} - 50}{x ^{2} - 25}

Maximum (0, - 2)

Solution steps

f^{'} (x) = - \frac{50 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} - 50}{x ^{2} - 25} : - 2

Maximum (0, - 2)

f (x) = (\frac{1}{8}) e^{- x}

U (x, y) = - x^{2} - y^{2} + 22 x + 18 y - 102

f (x) = - 2 x + 7

f (t) = 7 e^{- 7 t} u (t)

y = (x - \frac{200}{x})

What is the extreme f(x)=-(x^2-50)/(x^2-25) ?
The extreme f(x)=-(x^2-50)/(x^2-25) is Maximum(0,-2)