What is the extreme (16x)/(x^2+4) ?

The extreme (16x)/(x^2+4) is Minimum(-2,-4),Maximum(2,4)

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extreme (16x)/(x^2+4)

extreme points

\frac{16 x}{x ^{2} + 4}

Minimum (- 2, - 4), Maximum (2, 4)

Solution steps

f^{'} (x) = \frac{16 ( - 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{16 x}{x ^{2} + 4} : - 4

Minimum (- 2, - 4)

Plug

x = 2

into

\frac{16 x}{x ^{2} + 4} : 4

Maximum (2, 4)

Minimum (- 2, - 4), Maximum (2, 4)

f (x) = x^{2} - 4 x - 5

y = \frac{x}{x + 1}

(- 1, 2), (1, - 2)

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

What is the extreme (16x)/(x^2+4) ?
The extreme (16x)/(x^2+4) is Minimum(-2,-4),Maximum(2,4)