# How do you find the domain and range of g(t) = sqrt(3-t) - sqrt(2+t)?

May 17, 2018

#### Answer:

Domain is $- 2 \le t \le 3$ i.e. $\left[- 2 , 3\right]$ and range is $- \sqrt{5} \le g \left(t\right) \le \sqrt{5}$ i.e. $\left[- \sqrt{5} , \sqrt{5}\right]$

#### Explanation:

In the function $g \left(t\right) = \sqrt{3 - t} - \sqrt{2 + t}$

we can only have a non-negative number under square root sign.

Hence we ought to have $3 - t \ge 0$ i.e. $t \le 3$ and $2 + t \ge 0$ i.e. $t \ge - 2$

and hence domain is $- 2 \le t \le 3$

As $t$ takes value $- 2$, $g \left(t\right) = \sqrt{5}$ and as $t$ takes value $3$ $g \left(t\right) = - \sqrt{5}$

Hence range is $- \sqrt{5} \le g \left(t\right) \le \sqrt{5}$

graph{sqrt(3-x)-sqrt(2+x) [-6, 6, -3, 3]}