What is the extreme f(x)=(-x)/(x^2+7) ?

The extreme f(x)=(-x)/(x^2+7) is Maximum(-sqrt(7), 1/(2sqrt(7))),Minimum(sqrt(7),-1/(2sqrt(7)))

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extreme f(x)=(-x)/(x^2+7)

extreme points

f (x) = \frac{- x}{x ^{2} + 7}

Maximum (- 7, \frac{1}{2 7}), Minimum (7, - \frac{1}{2 7})

Solution steps

f^{'} (x) = - \frac{- x ^{2} + 7}{( x ^{2} + 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 7,

Decreasing

: - 7 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 7

into

\frac{- x}{x ^{2} + 7} : \frac{1}{2 7}

Maximum (- 7, \frac{1}{2 7})

Plug

x = 7

into

\frac{- x}{x ^{2} + 7} : - \frac{1}{2 7}

Minimum (7, - \frac{1}{2 7})

Maximum (- 7, \frac{1}{2 7}), Minimum (7, - \frac{1}{2 7})

\frac{x ^{2} + x - 12}{x ^{2} + x}

f (x) = \frac{9}{x + 2}

f (x) = x - 3 - 3

f (x) = lo g_{3} (x - 1)

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

What is the extreme f(x)=(-x)/(x^2+7) ?
The extreme f(x)=(-x)/(x^2+7) is Maximum(-sqrt(7), 1/(2sqrt(7))),Minimum(sqrt(7),-1/(2sqrt(7)))