What is the range of f(x)=\sqrt[3]{x+1}+5 ?

The range of f(x)=\sqrt[3]{x+1}+5 is -infinity <f(x)<infinity

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range of f(x)=\sqrt[3]{x+1}+5

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} - 15 x^{2} + 75 x - 126

Find the domain of each inverse function

Domain of

x^{3} - 15 x^{2} + 75 x - 126 : - \infty < x < \infty

Combine the intervals

- \infty < f (x) < \infty

(- 2.5) (10)

(- 1.4) (6.1)

f (x) = 8 x^{3} - 10

f (x) = 1 + \frac{9}{x}

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

What is the range of f(x)=\sqrt[3]{x+1}+5 ?
The range of f(x)=\sqrt[3]{x+1}+5 is -infinity <f(x)<infinity