What is the extreme f(x)=(-4x)/(x^2+3) ?

The extreme f(x)=(-4x)/(x^2+3) is Maximum(-sqrt(3), 2/(sqrt(3))),Minimum(sqrt(3),-2/(sqrt(3)))

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extreme f(x)=(-4x)/(x^2+3)

extreme

Maximum (- 3, \frac{2}{3}), Minimum (3, - \frac{2}{3})

Solution steps

f^{'} (x) = - \frac{4 ( - x ^{2} + 3 )}{( x ^{2} + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{- 4 x}{x ^{2} + 3} : \frac{2}{3}

Maximum (- 3, \frac{2}{3})

Plug

x = 3

into

\frac{- 4 x}{x ^{2} + 3} : - \frac{2}{3}

Minimum (3, - \frac{2}{3})

Maximum (- 3, \frac{2}{3}), Minimum (3, - \frac{2}{3})

What is the extreme f(x)=(-4x)/(x^2+3) ?
The extreme f(x)=(-4x)/(x^2+3) is Maximum(-sqrt(3), 2/(sqrt(3))),Minimum(sqrt(3),-2/(sqrt(3)))