What is the extreme (24x)/(x^2+16) ?

The extreme (24x)/(x^2+16) is Minimum(-4,-3),Maximum(4,3)

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extreme (24x)/(x^2+16)

extreme points

\frac{24 x}{x ^{2} + 16}

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

Solution steps

f^{'} (x) = \frac{24 ( - x ^{2} + 16 )}{( x ^{2} + 16 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 4

into

\frac{24 x}{x ^{2} + 16} : - 3

Minimum (- 4, - 3)

Plug

x = 4

into

\frac{24 x}{x ^{2} + 16} : 3

Maximum (4, 3)

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

y = - lo g_{4} (x)

(\frac{2}{3})^{x}

f (x) = \frac{24 t}{t ^{2} + 16}

(- 3, 5), (2, 6)

What is the extreme (24x)/(x^2+16) ?
The extreme (24x)/(x^2+16) is Minimum(-4,-3),Maximum(4,3)