### MODERATE EXERCISES-Square Roots and Other Radicals

(2) Find the domain of each function.

SOLUTION:

1. To determine the domain of this function, note that in order to take the square root, x+5 must be a non-negative number. So we set x+5 0 and solve this inequality.

Thus, the domain of is { x | x -5 }.

2. Similarly, in order to take the 4th root in the function g(t), we must have
.

To solve this inequality, we factor to get:

Then find the roots of the left-hand side:
t + 12 = 0 ==> t = -12
t - 5 = 0 ==> t = 5

These are the numbers at which the factors ( t + 12 ) and ( t - 5 ) can change sign.

If you think about it, you will see that when t < -12, the factor ( t + 12 ) is negative, and the factor ( t - 5 ) is also negative. So the product (t + 12)(t - 5) is positive when t < -12.

When -12 < t < 5, the factor (t + 12) is positive, and the factor ( t - 5 ) is negative. So the product (t + 12)(t - 5) is negative when -12 < t < 5.

These results, plus the case when t > 5, is summarized in the so-called "sign chart" below.

```
t+12          - - -          + + +          + + +
t-5           - - -          - - -          + + +
_______________o________________o___________________
-12               +5
(t+12)(t-5)       + + +          - - -          + + +
```

So (t + 12)(t - 5) is positive when t < -12 or when t > 5. And (t + 12)(t - 5)=0 when t = -12 or t = 5, so the solution to

is
{ t | t -12 or t 5 }.

This set is then the domain of g(t).