What is the extreme f(x)=-(8x)/(x^2+4) ?

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

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extreme f(x)=-(8x)/(x^2+4)

extreme points

f (x) = - \frac{8 x}{x ^{2} + 4}

Maximum (- 2, 2), Minimum (2, - 2)

Solution steps

f^{'} (x) = - \frac{8 ( - x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

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

Maximum (- 2, 2)

Plug

x = 2

into

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

Minimum (2, - 2)

Maximum (- 2, 2), Minimum (2, - 2)

f (x) = 3 x^{2} - x - 5

x^{2} + 52

\frac{2 - x}{1 - x}

6 x^{2} + 12 x - 1

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