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

- \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:

Inverse of

x^{3} - 3 x^{2} + 3 x - 1

Find the domain of each inverse function

Domain of

x^{3} - 3 x^{2} + 3 x - 1 : - \infty < x < \infty

Combine the intervals

- \infty < f (x) < \infty

f (x) = 6 x + 2

\frac{x - 7}{x ^{2} - 12 x + 35}

- x^{2} - 2 x + 8

16 x^{2} + 25 y^{2} = 100

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

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