What is the extreme f(x)= x/(x^2+6) ?

The extreme f(x)= x/(x^2+6) is Minimum(-sqrt(6),-1/(2sqrt(6))),Maximum(sqrt(6), 1/(2sqrt(6)))

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extreme f(x)= x/(x^2+6)

extreme points

f (x) = \frac{x}{x ^{2} + 6}

Minimum (- 6, - \frac{1}{2 6}), Maximum (6, \frac{1}{2 6})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 6}{( x ^{2} + 6 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = - 6

into

\frac{x}{x ^{2} + 6} : - \frac{1}{2 6}

Minimum (- 6, - \frac{1}{2 6})

Plug

x = 6

into

\frac{x}{x ^{2} + 6} : \frac{1}{2 6}

Maximum (6, \frac{1}{2 6})

Minimum (- 6, - \frac{1}{2 6}), Maximum (6, \frac{1}{2 6})

f (x) = ∣ x ∣ - 1

2 y = 4 x^{2} - 5

f (x) = 3^{x + 2}

(x - 4)^{2}

4 x + 3 y = 7, (- 2, - 9)

What is the extreme f(x)= x/(x^2+6) ?
The extreme f(x)= x/(x^2+6) is Minimum(-sqrt(6),-1/(2sqrt(6))),Maximum(sqrt(6), 1/(2sqrt(6)))