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### Section 1-1 : Review : Functions

29. Find the domain of $$f\left( z \right) = \sqrt {z - 1} + \sqrt {z + 6}$$.

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Hint : The domain of this function will be the set of all values of $$z$$ that will work in both terms of this function.
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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,

$z + 6 \ge 0\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}z \ge - 6$
Hint : What values of $$z$$ are in both of these?
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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,

${\mbox{Domain : }}z \ge 1$