Question 1099212
<br>Please note the other tutor's question about whether grouping symbols should be used.  The way you submitted the problem, the function is
{{{f(x) = x-3/x+2}}}<br>
I suspect that the actual function is the one she analyzed for you:
{{{f(x) = (x-3)/(x+2)}}}<br>
Then, to add to what she asked at the end of her response: Is the function value ever equal to 1?<br>
To answer that question, simply try solving the equation f(x) = 1.
{{{(x-3)/(x+2) = 1}}}
{{{x-3 = x+2}}}   [multiplying both sides by (x+2)]
{{{-3 = 2}}}<br>
There is no solution to this equation; the answer is no.  The function value is never equal to 1.<br>
It does in fact get as close as you want to 1, without ever reaching 1.  So the range of the function is all values except 1.<br>
So statements a and c are true.<br>
Statement b is false; it is easy to evaluate the function at x=3:
{{{f(3) = (3-3)/(3+2) = 0/5 = 0}}}<br>
Statement d is also false.  The only value not in the range of the function is 1.
You can also verify that statement d is false by solving the equation
{{{(x-3)/(x+2) = 2/3}}}
and finding that there is a solution.