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

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

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

range

y = \frac{1}{x ^{2} + 1}

0 < f (x) \leq 1

Interval Notation

(0, 1]

Solution steps

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

Domain of

\frac{1}{x ^{2} + 1} : - \infty < x < \infty

Extreme Points of

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

Maximum

(0, 1)

Find the range for the interval

- \infty < x < \infty : 0 < f (x) \leq 1

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

0 < f (x) \leq 1

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

x^{3} - 5 x

f (x) = \frac{( 3 x - 1 )}{2 x + 5}

y = 3 x^{6} - x^{8}

4 x - 2 - 6

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