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

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

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

range

\frac{1}{e ^{x} + 1}

0 < f (x) < 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}{e ^{x} + 1} : - \infty < x < \infty

Extreme Points of

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

None

Find the range for the interval

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

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

0 < f (x) < 1

f (x) = 18 x^{4} - 108 x^{2}

y = \frac{4 x ^{2} - 5}{2 x ^{3} + x}

f (x) = \frac{- x ^{2} + 25}{4 x + 20}

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