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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, - 2)

Plug

x = 2

into

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

Maximum (2, 2)

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

- 3 x^{5} + 5 x^{3} + 1

p (x) = 3 x^{3} - 10 x^{2} + a x - 2

f (x) = \frac{x ^{3} - 4}{x - 1}

f (x) = - \frac{2}{3} x^{3} + 10 x^{2} - 48 x - 1

f (x) = 4 x - 48 x^{\frac{1}{3}}

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