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Thread: finding the range

  1. #1
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    finding the range

    Hi folks,

    I really hope there is no simple arithmetic error in my answer, but I've checked it many times, so here goes.

    Show that for real x the function $y = \frac{x + 2}{(x + 3)(x - 1)}$ can take all real values.

    This should be really easy:

    $y = \frac{x + 2}{x^2 + 2x - 3}$

    $yx^2 + 2yx - 3y = x + 2 $

    $yx^2 + (2y - 1)x - 3y - 2 = 0$

    a quadratic in x of the form $ax^2 + bx + c = 0$. For real x the discriminant $(b^2 - 4ac \ge 0)$

    $(2y - 1)^2 + 4.y.(3y + 2) \ge 0$

    $4y^2 - 4y + 1 + 12y^2 + 8y \ge 0$

    $16y^2 + 4y + 1 \ge 0$

    solving for y, this has no real roots. I was expecting to get any expression like $y^2 + something \ge 0$ which would be true for all y, but no real roots suggests I have taken a wrong turn. Can you see it.
    Last edited by s_ingram; Jul 21st 2017 at 11:21 AM.
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  2. #2
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    Re: finding the range

    For any $y \in \mathbb{R}$:

    $x = \dfrac{1-2y \pm \sqrt{(2y-1)^2+4y(3y+2)}}{2y}$

    This is undefined for $y=0$. The discriminant is always positive for all $y$, so all that is left to check is that you can yet $y=0$, and it is trivial to verify that this occurs at $x=-2$. For any other value of $y$, you can find two values for $x$ such that when plugged in, gives you the desired $y$-value.
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  3. #3
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    Re: finding the range

    Thanks SlipEternal. I understand what you are saying, but I was puzzled by the terms inside the radical and why I could not solve the quadratic in y. What you have shown is that it doesn't matter. I just wondered if it had any significance.
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  4. #4
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    Re: finding the range

    Quote Originally Posted by s_ingram View Post
    Hi folks,

    I really hope there is no simple arithmetic error in my answer, but I've checked it many times, so here goes.

    Show that for real x the function $y = \frac{x + 2}{(x + 3)(x - 1)}$ can take all real values.
    Quote Originally Posted by s_ingram View Post
    I was puzzled by the terms inside the radical and why I could not solve the quadratic in y. What you have shown is that it doesn't matter. I just wondered if it had any significance.
    Frankly I am puzzled by the approaches period. You are not asked to find the inverse. So why?

    Let $y(x)=\dfrac{x+2}{(x+3)(x-1)}$, here is its graph.s

    Even without the graph it is easy to see that the only zero is $x=-2$.
    It is clear the function $y$ is continuous on $(-3,1)$. That important because:
    $\displaystyle {\lim _{x \to - {3^ + }}}y(x) = \infty \quad \& \quad {\lim _{x \to {1^ - }}}y(x) = - \infty $.
    That fact with continuity tells you that $y(x)$ is a surjuction to $\mathbb{R}$
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  5. #5
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    Re: finding the range

    Hi Plato,

    sketching the graph was the second part of the question. The idea is that we don't have an on-line graphing calculator and the task is first to establish the range of the function to help with the sketch. In similar questions in this area, the discriminant allows us to establish a function in y that is always greater than zero for real x. For example, if the function were $\dfrac{12}{(x + 3)(x - 1)}$ instead of $\dfrac{x+2}{(x + 3)(x - 1)}$ the range of the function would be $y \le -3, y \ge 0$ and these lines become like asymptotes on the graph. In the example of this post, the function in y $(16y^2 + 4y + 1 \ge 0)$ doesn't have real roots $y = -4 \pm \frac{\sqrt{16 - 4.16.1}}{32}$ my question on the post was how to interpret this result. I am expected to show that my function can take all real values, but I don't seem to have shown this, that's the first problem. The second is: What have I shown?
    Last edited by s_ingram; Jul 22nd 2017 at 05:52 AM.
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