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### Section 1-1 : Review : Functions

32. Find the domain of $$Q\left( y \right) = \sqrt {{y^2} + 1} - \sqrt[3]{{1 - y}}$$.

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The domain of this function will be the set of $$y$$’s that will work in both terms of this function. So, we need the domain of each of the terms.

For the first term let’s note that,

${y^2} + 1 \ge 1$

and so will always be positive. The domain of the first term is then all real numbers.

For the second term we need to notice that we’re dealing with the cube root in this case and we can plug all real numbers into a cube root and so the domain of this term is again all real numbers.

So, the domain of both terms is all real numbers and so the domain of the function as a whole must also be all real numbers or,

${\mbox{Domain : }} - \infty < y < \infty$