Section 1.1 : Review : Functions
20. Find the domain and range of \(f\left( z \right) = 2 + \sqrt {{z^2} + 1} \).
Show SolutionWe know that when we have square roots that we can’t take the square root of a negative number. However, because,
\[{z^2} + 1 \ge 1\]we will never be taking the square root of a negative number in this case and so the domain is all real numbers or,
\[{\mbox{Domain : }} - \infty < z < \infty \,\,\,{\rm{or}}\,\,\,\left( { - \infty ,\infty } \right)\]For the range we need to recall that square roots will only return values that are positive or zero and in fact the only way we can get zero out of a square root will be if we take the square root of zero. For our function, as we’ve already noted, the quantity that is under the root is always at least 1 and so this root will never be zero. Also recall that we have the following fact about square roots,
\[{\mbox{If }}x \ge 1{\mbox{ then }}\sqrt x \ge 1\]So, we now know that,
\[\sqrt {{z^2} + 1} \ge 1\]Finally, we are adding 2 onto the root and so we know that the function must always be greater than or equal to 3 and so the range is,
\[{\mbox{Range : }}\left[ {3,\infty } \right)\]