What is the extreme f(x)=(sqrt(x))/(1+x^2) ?

The extreme f(x)=(sqrt(x))/(1+x^2) is Minimum(0,0),Maximum(1/(sqrt(3)),(3^{3/4})/4)

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extreme f(x)=(sqrt(x))/(1+x^2)

extreme points

f (x) = \frac{x}{1 + x ^{2}}

Minimum (0, 0), Maximum (\frac{1}{3}, \frac{3 ^{\frac{3}{4}}}{4})

Solution steps

f^{'} (x) = \frac{1 - 3 x ^{2}}{2 x ( 1 + x ^{2} ) ^{2}}

Find intervals:

Increasing

: 0 < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < \infty

Plug

x = 0

into

\frac{x}{1 + x ^{2}} : 0

Minimum (0, 0)

Plug

x = \frac{1}{3}

into

\frac{x}{1 + x ^{2}} : \frac{3 ^{\frac{3}{4}}}{4}

Maximum (\frac{1}{3}, \frac{3 ^{\frac{3}{4}}}{4})

Minimum (0, 0), Maximum (\frac{1}{3}, \frac{3 ^{\frac{3}{4}}}{4})

What is the extreme f(x)=(sqrt(x))/(1+x^2) ?
The extreme f(x)=(sqrt(x))/(1+x^2) is Minimum(0,0),Maximum(1/(sqrt(3)),(3^{3/4})/4)