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

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

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

extreme

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

Solution steps

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

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

Plug

x = 3

into

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

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

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

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