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

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Answer 1 · 2018-05-17T06:01:24

Domain is $-2<=t<=3$ i.e. $[-2,3]$ and range is $-sqrt5<=g(t)<=sqrt5$ i.e. $[-sqrt5,sqrt5]$

In the function $g(t)=sqrt(3-t)-sqrt(2+t)$

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

Hence we ought to have $3-t>=0$ i.e. $t<=3$ and $2+t>=0$ i.e. $t>=-2$

and hence domain is $-2<=t<=3$

As $t$ takes value $-2$ , $g(t)=sqrt5$ and as $t$ takes value $3$ $g(t)=-sqrt5$

Hence range is $-sqrt5<=g(t)<=sqrt5$

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

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