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

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

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

range

\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

- 6 sin (3 x + \frac{π}{2})

x^{2} - 2 x - 8

f (x) = 16 x^{5} - 12 x^{3} + 4 x^{2} - 3

f (x) = \frac{x}{8} - 3

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

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