# Thread: Reciprocal functions - intersecting points

1. ## Reciprocal functions - intersecting points

Hi,

I hope someone can help. I had to graph the reciprocal of the function f(x) = x^2-3x+2. The solution is as follows:

I'm trying to understand why the red function (in the image attached) is so low in the centre. I'm referring to the point around (-1.5, -4). When I tried to find these points algebraically, I just got imaginary solutions. Could someone explain why the textbook provided this solution?

Sincerely,
Olivia

2. ## Re: Reciprocal functions - intersecting points

$x^2-3x+2 = (x-2)(x-1)$

so the reciprocal is

$\dfrac{1}{(x-2)(x-1)}$

there reciprocal will blow up to $\pm \infty$ at either of those poles, $x=2,~x=1$

by inspection

$\dfrac{1}{f(x)} = \begin{cases} + &x<1 \\ - &1<x<2 \\ + &2<x \end{cases}$

$f\left(\dfrac 3 2\right) = \dfrac 9 4 - \dfrac 9 2 + 2 = -\dfrac 1 4$

so $\dfrac{1}{f(3/2)} = -4$

maybe the error comes from you using $-\dfrac 3 2$ instead of $\dfrac 3 2$

3. ## Re: Reciprocal functions - intersecting points

$f(x) = x^2-3x+2 = (x-1)(x-2)$

If you plug in x=1 or x=2, you get 0. If you take the reciprocal, you have:

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

If you plug in 1 or 2 for x, you get a zero in the denominator. As you know, division by zero is not a defined operation, so the reciprocal function $g(x)$ is not defined for $x=1$ or $x=2$. Now, if $x<1$, then $x-1<0$ and $x-2<0$. When you multiply two negative numbers together, you get a positive number. When $1<x<2$, then $x-1>0$, but $x-2<0$. So, you are multiplying a positive and a negative number (which is why the graph is below the x-axis between $x=1$ and $x=2$). If $x>2$, then both $x-1>0$ and $x-2>0$, so you are multiplying two positive numbers which is why the function becomes positive again.

Now, what happens when you are VERY close to the points $x=1$ and $x=2$? Let's say $x=0.999$. Then, $x-1 = 0.001$ and $x-2 = 1.001$. When you multiply these two numbers together, you get 0.001001, which is a very small number. But, if you divide 1 by a very small number, you get a very big number. In fact, as you get closer and closer to 1: $x=0.999, x=0.9999, x=0.99999, \ldots$, you are dividing 1 by smaller and smaller numbers, which produces larger and larger results. That is why we say the function approaches infinity as $x$ approaches 1 from the left. Now, as $x$ approaches 1 from the right, we have a positive number times a negative number, so we wind up with a negative number, but it is a negative number of very small magnitude. When you divide positive 1 by a negative number of very small magnitude, you get a negative number of very big magnitude. So we say that as $x$ approaches 1 from the right, the function approaches negative infinity. You can never reach infinity or negative infinity. You can only approach it. That is why the function looks strange around those two points.