What is the domain and range of #f(x) =(3x^2-2x-8)/(2x^3+x^2-3x)#?

1 Answer
Apr 27, 2017

Answer:

Domain: #(-\infty,-3/2)\cup(-3/2,0)\cup(0,1)\cup(1,\infty)#
Range: #(-\infty,\infty)#

Explanation:

To find the domain, we have to look for any cases where division by zero can occur. In this case, we have to make sure #2x^3+x^2-3x\ne 0# To solve this we can simplify by factoring out an #x#.
#x(2x^2+x-3)\ne 0#

Solving we have two options
#x\ne 0# and #2x^2+x-3\ne 0#

We have to solve the second equation to get
#\frac{-(1)\pm\sqrt{(1)^2-4(2)(-3)}}{2(2)}#
#\frac{-1\pm\sqrt{1+24}}{4}#
#\frac{-1\pm 5}{4}#
#\frac{-1+5}{4}=4/4=1#
#\frac{-1-5}{4}=-6/4=-3/2#

So the function is undefined at #x=-3/2,0,1#
This means our domain is
#(-\infty,-3/2)\cup(-3/2,0)\cup(0,1)\cup(1,\infty)#

As you get closer to any of those x-values we found, the denominator gets closer to 0. As the denominator gets closer to 0, the resulting value goes to positive or negative infinity so the range is #(-\infty,\infty)#.