What is the extreme f(x)=(5x)/(x^2+2) ?

The extreme f(x)=(5x)/(x^2+2) is Minimum(-sqrt(2),-5/(2sqrt(2))),Maximum(sqrt(2), 5/(2sqrt(2)))

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extreme f(x)=(5x)/(x^2+2)

extreme points

f (x) = \frac{5 x}{x ^{2} + 2}

Minimum (- 2, - \frac{5}{2 2}), Maximum (2, \frac{5}{2 2})

Solution steps

f^{'} (x) = \frac{5 ( - x ^{2} + 2 )}{( x ^{2} + 2 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{5 x}{x ^{2} + 2} : - \frac{5}{2 2}

Minimum (- 2, - \frac{5}{2 2})

Plug

x = 2

into

\frac{5 x}{x ^{2} + 2} : \frac{5}{2 2}

Maximum (2, \frac{5}{2 2})

Minimum (- 2, - \frac{5}{2 2}), Maximum (2, \frac{5}{2 2})

y = \frac{x}{2} - sin (x)

\frac{x - 5}{x ^{2} - 16 x + 64}

f (x) = 3 x^{\frac{2}{3}} - x, 0 \leq x \leq 27

f (x) = \frac{5 x}{x ^{2} + 9}

x^{9} - 9 x

What is the extreme f(x)=(5x)/(x^2+2) ?
The extreme f(x)=(5x)/(x^2+2) is Minimum(-sqrt(2),-5/(2sqrt(2))),Maximum(sqrt(2), 5/(2sqrt(2)))