What is the range of sqrt((4x)/(x^2+1)) ?

The range of sqrt((4x)/(x^2+1)) is 0<= f(x)<= sqrt(2)

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range of sqrt((4x)/(x^2+1))

range

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

0 \leq f (x) \leq 2

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{4 x}{x ^{2} + 1} : x \geq 0

Extreme Points of

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

Minimum

(0, 0),

Maximum

(1, 2)

Find the range for the interval

0 \leq x < \infty : 0 \leq f (x) \leq 2

Combine the ranges of all domain intervals to obtain the function range

0 \leq f (x) \leq 2

f (x) = \frac{x - 1}{4}

f (x) = - \frac{1}{x + 5}

f (x) = 9 (\frac{2}{x}) - 4

4 x^{2} + 7

\frac{5 + x}{4 - 2 x}

What is the range of sqrt((4x)/(x^2+1)) ?
The range of sqrt((4x)/(x^2+1)) is 0<= f(x)<= sqrt(2)