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I am wondering how many intervals would I have to set up the chart. I keep on getting questions such as this wrong due to the fact that I don't evaluate the correct amount of intervals,

My intervals from $(4-x^2)/x$ are:

$(-\infty,2)$, $(-2,2)$ and $(2,\infty)$

However, in the answer key I see $(0,2)$ is also a interval, why? Also, why has the inequality sign not been inverted?

Please show work.

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  • $\begingroup$ Do you mean $\dfrac{4}{x}<x$? $\endgroup$
    – user122283
    Apr 8, 2014 at 2:41
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    $\begingroup$ In the title of the question you have $4/x < x$ but in the question itself you have $(4-x^2)/x$. which one are you trying to analyze? $\endgroup$
    – user139388
    Apr 8, 2014 at 2:41
  • $\begingroup$ @user139388 It seems like he is trying to solve $(4-x^2)/x=4/x-x<0$, though I'm not entirely sure. $\endgroup$
    – user122283
    Apr 8, 2014 at 2:41
  • $\begingroup$ okay I thought it was okay to multiply the left side by a negative; thus, changing the -x^2 term to a positive. But that was not done. thus the equation is (x-2)(x+2)/x <0. However, that still begs the question; the intervals would still be in (- infinity, -2) (-2,2) (2,infinity) Where is the (2,0) and (0,2)interval coming from? $\endgroup$
    – Prologue
    Apr 8, 2014 at 2:49

2 Answers 2

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The way to work with such inequalities is to move everything to one side and compare some expression to zero. In this case, you would have $$x-\frac4x>0.$$ The method then is to find where the expression is equal to zero, or undefined; these points then break up the real line into intervals on which the expression has constant sign, and you just pick the ones with the sign you want.

That usually means you must factor the expression. We proceed as follows: $$x-\frac4x>0$$ $$\frac{x^2-4}{x}>0$$ $$\frac{(x+2)(x-2)}{x}>0$$ The numerator is zero if $x=-2$ or $x=2$. (These points are where the whole expression is zero.)

The denominator is zero if $x=0$. (This point is where the whole expression is undefined.)

So, the points of interest are, in order from left to right, $-2$, $0$, and $2$. The intervals that they determine (that is, what's left when they are removed) are $(-\infty,-2)$, $(-2,0)$, $(0,2)$, and $(2,\infty)$.

Perhaps the easiest way to determine the sign of the whole expression on each of these intervals is to find the sign of each factor and multiply signs.

$(x+2)$ is negative to the left of $-2$ and positive to the right of $-2$.

$(x-2)$ is negative to the left of $2$ and positive to the right of $2$.

$x$ is negative to the left of $0$ and positive to the right of $0$.

So:

$\frac{(x+2)(x-2)}{x}$ has sign $(-)(-)(-)=-$ on the interval $(-\infty,-2)$.

$\frac{(x+2)(x-2)}{x}$ has sign $(+)(-)(-)=+$ on the interval $(-2,0)$.

$\frac{(x+2)(x-2)}{x}$ has sign $(+)(-)(+)=-$ on the interval $(0,2)$.

$\frac{(x+2)(x-2)}{x}$ has sign $(+)(+)(+)=+$ on the interval $(2,\infty)$.

So, putting it together, $\frac{(x+2)(x-2)}{x}>0$ if $x$ is in either $(-2,0)$ or $(2, \infty)$. That is, the solution set is $$\boxed{(-2,0)\cup(2,\infty)}.$$

Note that if we were looking for "$\geq$" instead of "$>$", we would just include the interval endpoints that make the numerator zero.

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  • $\begingroup$ Thanks so much for the thorough explanation. $\endgroup$
    – Prologue
    Apr 8, 2014 at 3:39
  • $\begingroup$ You're welcome. This is the only tractable way to solve such inequalities. You can't multiply by a variable expression without imposing complicated conditions on the resulting inequalities. This avoids the mess and is really easy to understand. $\endgroup$
    – MPW
    Apr 8, 2014 at 3:44
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$$\dfrac{4}{x}<x\\ \implies 4<x^2\text{ if $x>0$, else }4>x^2\\ \implies x\in(-\infty,-2)\cup(2,\infty)\cup(-2,2)\setminus\{0\}$$ To see why $(0,2)$ was included, see that $(0,2)\subset(-2,2)\setminus\{0\}$.

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  • $\begingroup$ Please look over my last question. $\endgroup$
    – Prologue
    Apr 8, 2014 at 2:50
  • $\begingroup$ @Sanath, in your first step you have multiplied the inequality by a number which might be negative. $\endgroup$
    – David
    Apr 8, 2014 at 2:53
  • $\begingroup$ @David Ah right! That's where I went wrong. Thanks. See my edited answer. $\endgroup$
    – user122283
    Apr 8, 2014 at 2:55
  • $\begingroup$ I think you are misunderstanding my question. When I set up the "sign" diagram (to test which parts of the real number line are positive or negative). I have the intervals (- infinity, -2) (-2,2) (2,infinity) Where is the (2,0) and (0,2)interval coming from? $\endgroup$
    – Prologue
    Apr 8, 2014 at 2:56
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    $\begingroup$ @Prologue The inequality $4>x^2$ is saturated iff $x\in(-2,2)$. However, for the original inequality, $4/x<x$, $x=0$ is not a solution. Hence, $(-2,2)\setminus\{0\}\supset(0,2)$ is included. $\endgroup$
    – user122283
    Apr 8, 2014 at 3:00

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