Finding the range of $f(x) = 1/((x-1)(x-2))$

I want to find the range of the following function

$$f(x) = \frac{1}{(x-1)(x-2)} $$

Is there any way to find the range of the above function? I have found one idea. But that is too critical. Please help me in solving this problem .

Solution 1:

HINT:

Let $$y=\frac1{(x-1)(x-2)}$$

$$\implies (y)x^2-(3y)x+2y-1=0$$ which is a Quadratic Equation in $x$

As $x$ is real, the discriminant must be $\ge 0$

$$\implies (3y)^2-4\cdot y\cdot(2y-1)\ge0$$

$$\implies y^2+4y\ge 0\iff y(y+4)\ge 0$$

Either $y\ge 0$ and $y+4\ge 0\implies y\ge -4\implies y\ge0$

or $y\le 0$ and $y+4\le 0\implies y\le -4\implies y\le-4$

Solution 2:

In addition to lab bhattacharjee's answer, I'd recommend you to think on the following (in this and other cases when you are after the range of the function)

1. For two values y is undefined, because you can't divide by 0

2. y is strictly less than 0 only for some small interval (which?), due to the definition of the denominator

3. For the same reason, for two open intervals(which?) y is strictly positive
4. You can conclude from these three statements that y never takes the value 0.

Solution 3:

You are probably familiar with the curve $y=(x-1)(x-2)$. It is an upward opening parabola that crosses the $x$ axis at $x=1$ and $x=2$, and reaches a minimum value of $-\frac{1}{4}$ midway between $x=1$ and $x=2$.

Draw a picture of the parabola.

Note the function $y$, and therefore also $\frac{1}{y}$, is symmetric about $x=\frac{3}{2}$. So for the range of $\frac{1}{y}$ we need only look at $x\ge \frac{3}{2}$.

As $x$ travels over the interval $(2,\infty)$, the qantity $\frac{1}{y}$ goes from very large positive values towards $0$. That contributes the interval $(0,\infty)$ to the range.

As $x$ travels from $x=\frac{3}{2}$ towards $x=2$, the quantity $\frac{1}{y}$ travels from $-4$ towards $0$, through negative values. That conttributes the interval $[-4,0)$ to the range.

So the full range is $[-4,0)\cup (0,\infty)$.