Question

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

Answer 1

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)`.