What is the range of f(x)=\sqrt[3]{x+2} ?

The range of f(x)=\sqrt[3]{x+2} is -infinity <f(x)<infinity

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range of f(x)=\sqrt[3]{x+2}

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} - 2

Find the domain of each inverse function

Domain of

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

Combine the intervals

- \infty < f (x) < \infty

x - \frac{5}{3}

f (x) = \frac{2 - x ^{2}}{x ^{2} + 2 x - 48}

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

e^{x}

f (x) = \frac{x}{x ^{2} + 17 x + 66}

What is the range of f(x)=\sqrt[3]{x+2} ?
The range of f(x)=\sqrt[3]{x+2} is -infinity <f(x)<infinity