# Thread: IROC rational function

1. ## IROC rational function

Hi,

I hope someone can help. I'm trying to find two points on f(x) = (x+2)/(x-1) where the IROC is -3. I created the difference quotient for the equation and now I just have to solve for x in order to figure out these two points where the IROC is -3.

REALLY what I need guidance on is the meaning behind the the solutions for x. For example, when I plug in 0.001 for h I get two solutions, and when I plug in -0.001 for h I get another two solutions for x. Does this mean I have total of 4 solutions for x on both ends of the instantaneous rate of change of -3? I hope someone can help clarify this confusion for me.

Here the examples I am referring to:
https://www.wolframalpha.com/input/?...1)(x%2B0.001-1))
https://www.wolframalpha.com/input/?...x-1)(x-0.001-1))

Also I don't know how to use derivatives yet, so I would appreciate a response which isn't too focused on that.

Sincerely,
Olivia

2. ## Re: IROC rational function

The IROC is approximately the slope of the line between two points on the curve. We have point 1: $(x,f(x))$ and point 2: $(x+h,f(x+h))$. Let's write the formula for the slope between these two lines:

$\dfrac{f(x+h)-f(x)}{(x+h)-x} = \dfrac{f(x+h)-f(x)}{h}$

Now, let's plug in the formula we have for $f(x)$:

$\dfrac{\dfrac{x+h+2}{x+h-1}-\dfrac{x+2}{x-1}}{h} = \dfrac{(x+h+2)(x-1)-(x+2)(x+h-1)}{h(x+h-1)(x-1)}$

Now, let's simplify:

$\dfrac{(x^2+hx+2x-x-h-2)-(x^2+hx-x+2x+2h-2)}{h(x+h-1)(x-1)} = \dfrac{-3h}{h(x+h-1)(x-1)} = \dfrac{-3}{(x+h-1)(x-1)}$

Now, if $h$ is really, really close to zero, this is very close to:

$\dfrac{-3}{(x-1)^2}$

So, let $\dfrac{-3}{(x-1)^2} = -3$

We get $(x-1)^2 = 1$, so $x-1 = \pm 1$, so $x=0, x=2$. Check out the IROC at those two values of $x$. Chances are, it will be really close to $-3$. In fact, in each of the examples you showed above, you got one value for $x$ very close to 0 and one very close to 2.

(Actually, it will be exactly $-3$, because this is basically the method you would use to find a derivative.)

3. ## Re: IROC rational function

Thank you your response! Are you suggesting that based on my examples above the solutions sets could either be (x = -0.000500125, and x = 1.9995) or (x = 0.000499875, and x = 2.0005)? I'm still a bit confused.

4. ## Re: IROC rational function

I am saying that the instantaneous rate of change at $x=0$ and at $x=2$ is $-3$. Your examples are good approximations of the actual answer. Since you are taking approximations, you could round them to $x=0$ and $x=2$.

5. ## Re: IROC rational function

Okay good to know Thank you very much.