Finding the range of $((x+1)(x-2))/(x-3)$ analytically

Question

I know the range of $f(x)=\frac{(x+1)(x-2)}{x-3}$ is $(-\infty,1]\cup[9,\infty)$, but is there an easy way of determining this analytically (i.e., without using a calculator)? I do not see any easy way of doing this.

Answer 1

The equation $$ \dfrac{(x+1)(x-2)}{x-3} = y$$ is equivalent to the quadratic $$x^2 - (y+1) x + 3 y - 2 = 0$$ assuming $x \ne 3$ (but $x=3$ is not a problem, because the quadratic equation is never true when $x=3$). That quadratic has real solutions if and only if its discriminant $y^2 - 10 y + 9 \ge 0$. Note that $y^2 - 10 y + 9 = 0$ for $y=1$ and $y=9$ and its graph is a parabola opening upwards, so $y^2 - 10 y + 9 \ge 0$ for $y \in (-\infty, 1] \cup [9, \infty)$.

Answer 2

You flagged this as a precalculus question, but assuming some knowledge of calculus, you could take the derivative, and search for critical points.

We know $f'(x)={x^2 - 6x + 5 \over{{(x-3)}^2}} $, so its critical points are at 1 and 5. At $1$, $f''(x)\lt0$, and at $5$, $f''(x)\gt0$, so these are local maxima and minima.

Then, by evaluating at these points, we get the domain.