What is the extreme f(x)=(3x)/(x^2+1) ?

The extreme f(x)=(3x)/(x^2+1) is Minimum(-1,-3/2),Maximum(1, 3/2)

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extreme f(x)=(3x)/(x^2+1)

extreme

Minimum (- 1, - \frac{3}{2}), Maximum (1, \frac{3}{2})

Solution steps

f^{'} (x) = \frac{3 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{3 x}{x ^{2} + 1} : - \frac{3}{2}

Minimum (- 1, - \frac{3}{2})

Plug

x = 1

into

\frac{3 x}{x ^{2} + 1} : \frac{3}{2}

Maximum (1, \frac{3}{2})

Minimum (- 1, - \frac{3}{2}), Maximum (1, \frac{3}{2})

What is the extreme f(x)=(3x)/(x^2+1) ?
The extreme f(x)=(3x)/(x^2+1) is Minimum(-1,-3/2),Maximum(1, 3/2)