I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 1.1 : Review : Functions
32. Find the domain of \(Q\left( y \right) = \sqrt {{y^2} + 1} - \sqrt[3]{{1 - y}}\).
Show SolutionThe 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,
\[{\mbox{Domain : }} - \infty < y < \infty \]