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

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

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

\frac{x ^{3} - 1}{27} + \frac{- x ^{2} + x}{9} + 3

Find the domain of each inverse function

Domain of

\frac{x ^{3} - 1}{27} + \frac{- x ^{2} + x}{9} + 3 : - \infty < x < \infty

Combine the intervals

- \infty < f (x) < \infty

f (x) = x - 8

f (x) = 9 x - 10

f (x) = \frac{( x - 3 )}{x ^{2} - 4}

f (x) = 4 x^{2} - 12 x + 9

20 x csc (5 x^{2} + 2) d x

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