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

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

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

range

f (x) = \frac{4 x}{x ^{2} + 1}

0 \leq f (x) \leq 2

Interval Notation

[0, 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 ^{2}}{x ^{2} + 9}

(- 1.13) (- 4 - 5)

\frac{1}{x + 1}

\frac{10}{x}

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

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