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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

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

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

Plug

x = 1

into

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

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

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

x^{\frac{1}{9}} + 1

f (x) = \frac{3 x}{x - 2}

f (x, y) = x y + \frac{8}{x} + \frac{8}{y}

f (x, y) = x^{3} + y^{3} - 300 x - 75 y - 3

f (x) = x^{2} + \frac{12}{x}

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