What is the range of f(x)=\sqrt[3]{x+1} ?

The range of f(x)=\sqrt[3]{x+1} is -infinity <f(x)<infinity

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range of f(x)=\sqrt[3]{x+1}

range

f (x) = 3 x + 1

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

3 x + 1

Inverse of

3 x + 1 : x^{3} - 1

x^{3} - 1

Find the domain of each inverse function

Domain of

x^{3} - 1 : - \infty < x < \infty

Combine the intervals

- \infty < f (x) < \infty

f (x) = \frac{1}{3} x + 6

y = \frac{7}{1 + t}

5 sin (2 x + π)

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

y = 2.5 x + 7

What is the range of f(x)=\sqrt[3]{x+1} ?
The range of f(x)=\sqrt[3]{x+1} is -infinity <f(x)<infinity