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Thread: IROC rational function

  1. #1
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    Question 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
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  2. #2
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    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.)
    Last edited by SlipEternal; May 8th 2017 at 09:10 AM.
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  3. #3
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    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.
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  4. #4
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    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 $x=2$.
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  5. #5
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    Re: IROC rational function

    Okay good to know Thank you very much.
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