Section 1.1 : Review : Functions
29. Find the domain of \(f\left( z \right) = \sqrt {z - 1} + \sqrt {z + 6} \).
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The domain of this function will be the set of all \(z\)’s that we can plug into both terms in this function and get a real number back as a value. This means that we first need to determine the domain of each of the two terms.
For the first term we need to require,
\[z - 1 \ge 0\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}z \ge 1\]For the second term we need to require,
\[z + 6 \ge 0\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}z \ge - 6\]Now, we just need the set of \(z\)’s that are in both conditions above. In this case notice that all the \(z\) that satisfy \(z \ge 1\) will also satisfy \(z \ge - 6\). The reverse is not true however. Any \(z\) that is in the range \( - 6 \le z < 1\) will satisfy \(z \ge 6\) but will not satisfy \(z \ge 1\).
So, in this case, the domain is in fact just the first condition above or,
\[{\mbox{Domain : }}z \ge 1\]