What is the extreme f(x)=-(3x)/(x^2+8) ?

The extreme f(x)=-(3x)/(x^2+8) is Maximum(-2sqrt(2), 3/(4sqrt(2))),Minimum(2sqrt(2),-3/(4sqrt(2)))

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extreme f(x)=-(3x)/(x^2+8)

extreme points

f (x) = - \frac{3 x}{x ^{2} + 8}

Maximum (- 22, \frac{3}{4 2}), Minimum (22, - \frac{3}{4 2})

Solution steps

f^{'} (x) = - \frac{3 ( - x ^{2} + 8 )}{( x ^{2} + 8 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 22,

Decreasing

: - 22 < x < 22,

Increasing

: 22 < x < \infty

Plug

x = - 22

into

- \frac{3 x}{x ^{2} + 8} : \frac{3}{4 2}

Maximum (- 22, \frac{3}{4 2})

Plug

x = 22

into

- \frac{3 x}{x ^{2} + 8} : - \frac{3}{4 2}

Minimum (22, - \frac{3}{4 2})

Maximum (- 22, \frac{3}{4 2}), Minimum (22, - \frac{3}{4 2})

f (x, y) = 50 y^{2} + x^{2} - x^{2} y

f (x) = x lo g_{e} (x)

f (x) = 2 x + 5 y

f (x) = 2 x + 7 y

What is the extreme f(x)=-(3x)/(x^2+8) ?
The extreme f(x)=-(3x)/(x^2+8) is Maximum(-2sqrt(2), 3/(4sqrt(2))),Minimum(2sqrt(2),-3/(4sqrt(2)))