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

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

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

extreme points

f (x) = \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)

f (x) = ln (5 - x) + ln (x - 3)

f (x) = lo g_{3} (x - 6)

\frac{2 x ^{3}}{x ^{5}}

(x - 3) x

f (x) = 2 x^{3} - 27 x^{2} + 84 x

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