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

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The domain of the function ax^(1/3) + b is all real numbers, denoted as ℝ. This conclusion arises from the fact that the cube root function, x^(1/3), is defined for all real values of x. Consequently, regardless of the constants a and b, the function consistently produces real outputs for any real input. Therefore, the domain encompasses every real number.

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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.
 
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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.
 
Thanks. Good information.
 

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