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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = - 5

into

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

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

Plug

x = 5

into

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

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

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

f (x) = ln (x + 2) + \frac{4}{x + 2}

f (x) = 4 x^{7}

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

y = - 57.17 x^{2} + 621.83 x + 1089

f (x) = x + \frac{( 1 )}{( x )}, 0.2 \leq x \leq 4

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