Given a function:
y= $\frac{(2x^2+11x+15)^{1/2}}{(x^2-9)^{1/2}}$
What is the domain and range for this function?
My attempt:
To find domain, I need to find the value of x such that the function is defined:
$\ (x^2-9)$ >0
$\ (x-3)*(x+3)$ >0
So, we get first domain: x ≠ 3 and x ≠ -3
Another domain is:
$\frac{2x^2+11x+15}{x^2-9}$ >0
$\ (2x^2+11x+15)$ >0
$\ (x+3)(2x+3)$ >0
x= -3 or x= -3/2
At number line, we get the domain should be:
$\ x< -3 or x > -3/2 $, where x≠ 3 and x ≠ -3.
Is my domain true?
But, how to find the range? it seems pretty hard to find the range for this problem. Can somebody give me a clue or any hints to find the range?
Thanks
you have to solve two inequalities: $2x^2+11x+15\geq 0$ and $x^2-9\geq 0$ thus we have $x\le -3$ or $x\geq -\frac{5}{2}$ from the first inequality and $x<-3$ or $x>3$ from the second inequality.