What is the range of f(x)=-2\sqrt[3]{x-2}+1 ?

The range of f(x)=-2\sqrt[3]{x-2}+1 is -infinity <f(x)<infinity

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range of f(x)=-2\sqrt[3]{x-2}+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} + 3 x ^{2} - 3 x + 1}{8} + 2

Find the domain of each inverse function

Domain of

\frac{- x ^{3} + 3 x ^{2} - 3 x + 1}{8} + 2 : - \infty < x < \infty

Combine the intervals

- \infty < f (x) < \infty

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

f (x) = \frac{x ^{2} + x - 6}{x ^{2} + 6 x + 9}

f (x) = \frac{x ^{2} - 4 x}{x ^{2} - 16} x = 3.9

2 x - 4

f (x) = x^{2} - 2 x - 15

What is the range of f(x)=-2\sqrt[3]{x-2}+1 ?
The range of f(x)=-2\sqrt[3]{x-2}+1 is -infinity <f(x)<infinity