Why is the domain of ax^(1/3) + b equal to all real numbers?

[mathdad]

Find the domain.

ax^(1/3) + b

I understand that a and b are constants here. I also know that x^(1/3) is equivalent to cube root{x}.

What I do not understand is why the answer is ALL REAL NUMBERS.

 

[Greg]

It's "all real numbers" because no matter what real value you give x the function returns another real number. Given that a function's domain is defined wherever the function gives a real number for a real number input, ax^(/13) + b has the set $\mathbb{R}$ as its domain, i.e. it is defined for all real x.

 

[mathdad]

Thanks. Good information.