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

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

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

extreme

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

Solution steps

f^{'} (x) = - \frac{2 ( - 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{- 2 x}{x ^{2} + 7} : \frac{1}{7}

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

Plug

x = 7

into

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

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

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

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