What is the extreme f(x)=-(24x)/(x^2+9) ?

The extreme f(x)=-(24x)/(x^2+9) is Maximum(-3,4),Minimum(3,-4)

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extreme f(x)=-(24x)/(x^2+9)

extreme points

f (x) = - \frac{24 x}{x ^{2} + 9}

Maximum (- 3, 4), Minimum (3, - 4)

Solution steps

f^{'} (x) = - \frac{24 ( - x ^{2} + 9 )}{( x ^{2} + 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

- \frac{24 x}{x ^{2} + 9} : 4

Maximum (- 3, 4)

Plug

x = 3

into

- \frac{24 x}{x ^{2} + 9} : - 4

Minimum (3, - 4)

Maximum (- 3, 4), Minimum (3, - 4)

- 2 x^{2} - 12 x - 17

f (x) = 5 x e^{- x}, 0 \leq x \leq 2

f (x) = 2 x^{3} - 9 x^{2} + 60 x + 8

f (x) = 150 x - 2 x^{3}

f (x) = (x^{2} - 2 x - 8)^{\frac{1}{3}}

What is the extreme f(x)=-(24x)/(x^2+9) ?
The extreme f(x)=-(24x)/(x^2+9) is Maximum(-3,4),Minimum(3,-4)