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I'm stuck trying to solve an inequality of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$; specifically, $\left|\frac{2x + 5}{x + 1}\right| \geq 1$

I tried the approach that was taught for solving inequalities in general, and solved first for x when the part enclosed in absolute value signs is $\geq 1$, and then solved for when it is $\leq -1$, but plugging in numbers for the resulting domain $[-4, -2]$ back into the original inequality resulted in some answers that were not $\geq 1$.

Where have I gone wrong in trying to solve this inequality? How is this different from solving an inequality like $\left|\frac{3}{x + 1}\right| \geq 1$?

EDIT: Okay, since I'm clearly messing this up horribly(and now realize that I was absent-mindedly taking the intersection and not the union of my results in spite of having not mixed up the two many times before), I'll edit with the specifics of the solution I was trying and which had me confused before I asked the question. So:

$\left|\frac{2x + 5}{x + 1}\right| \geq 1$

Meaning 1: $\frac{2x + 5}{x + 1} \geq 1$ or 2: $\frac{2x + 5}{x + 1} \leq -1$

Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.

Now the other inequality: multiplied -1 by $x + 1$, again forgetting to consider $x + 1$ being negative: $2x + 5 \leq -x - 1$ , rearrange and divide by 3 for $x \leq -2$.

Domain is the union, so $[-4, \infty] \cup [-\infty, -2]$, so $R - { 1 }$. This is obviously not right, as plugging in -3 shows.

So, in inequalities where the denominator is some function g(x), if I split it and choose to multiply, will I have to solve each part of the inequality twice(once assuming g(x) is positive, a second time assuming it's negative) and then check which solution makes sense by plugging in numbers?

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Here are some thoughts for you! Can you simplify the quotient? What if you write $y=f/g$ and work with $|y|\ge1$? If $|2+z|\ge1$ what does this tell you about the distance between $z$ and $-2$? – AD. Sep 25 '10 at 21:19
up vote 6 down vote accepted

Your answer of $[-4,-2]$ appears to be the solution set to $\left|\frac{2x + 5}{x + 1}\right| \leq 1$. Without seeing all of your work, I can't be sure what happened.

In general, for inequalities of this type (actually, pretty much any inequalities involving a quotient with variables in the denominator), the method I'd suggest is the following:

  1. Split it into two inequalities to solve separately: $\frac{f(x)}{g(x)} \geq 1$ or $-1\geq \frac{f(x)}{g(x)}$. Once these are solved, take the union of their solution sets (only one of these inequalities must be true for your original inequality to hold).

  2. For each of these two inequalities, manipulate them so that the comparison is to 0 instead of 1 or -1. For example, $\frac{f(x)}{g(x)} \geq 1$ can be rewritten as $\frac{f(x)}{g(x)}-1\geq 0$ or $\frac{f(x)-g(x)}{g(x)} \geq 0$.

  3. When solving an inequality that compares an expression of the form $\frac{a(x)}{b(x)}$ to zero, the solution set will be the union of intervals whose endpoints are at the zeros of $a(x)$ or $b(x)$. That is:

    • determine the zeros of $a(x)$ and $b(x)$
    • plot the zeros on a number line
      • use open circles (excluded values) for the zeros of $b(x)$
      • if the inequality is $\leq$ of $\geq$, use closed circles (included values) for the zeros of $a(x)$
      • if the inequality is $<$ or $>$, use open circles for the zeros of $a(x)$
    • for each region between two zeros, test a value in the inequality you are solving
    • if that value makes the inequality true, then the region is part of the solution set

    This is sometimes called the "boundary algorithm" or the "test-point method" for solving inequalities.

edit: Regarding "if I split it and choose to multiply, will I have to solve each part of the inequality twice," well... sort of.

Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.

Here, you should have $(2x+5\geq x+1\text{ and }x\geq -1)\text{ or }(2x+5\leq x+1\text{ and }x<-1)$. In solving, remember that the "and" becomes an intersection of solution sets and the "or" becomes a union of solution sets.

As I'd indicated in the comments, I use the method I originally described in order to avoid this sort of casework.

edit 2: As an example of the method that I describe above, let's solve $\frac{2x+5}{x+1}\geq 1$. Subtract 1 from both sides, get a common denominator on the left side, and combine the terms to get $\frac{x+4}{x+1}\geq 0$. The only zero of the numerator is -4; the only zero of the denominator is -1. Graphing these on a number line gives: $$\leftarrow\!\!\underset{-4}{-\!\!\bullet\!\!-}\!\!-\!\!\underset{-1}{-\!\!\circ\!\!-}\!\!\rightarrow$$ (open circle at -1 because $x=-1$ makes the quotient undefined, closed circle at -4 because $x=-4$ makes the numerator 0, so the quotient is 0, and the inequality is $\geq 0$).

Now, there are 3 intervals to test: $(-\infty,-4]$, $[-4,-1)$, and $(-1,\infty)$. Relatively simple-to-test values in the intervals are -5, -2, and 0, respectively. $\frac{-5+4}{-5+1}=\frac{-}{-}=+\geq 0$; $\frac{-2+4}{-2+1}=\frac{+}{-}=-\not\geq 0$; $\frac{0+4}{0+1}=\frac{+}{+}=+\geq 0$. So, the solution is $(-\infty,-4]\cup(-1,\infty)$.

(To complete the original problem, you still need to solve $\frac{2x+5}{x+1}\leq -1$ and take the union of that solution with the set we just found.)

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The solution set of the inequality $$|f(x)/g(x)|\geq1$$ is equal to the union of the solutions sets of the inequalities $$f(x)/g(x)\leq-1$$ and $$f(x)/g(x)\geq1,$$ so solve these two and compute the union.

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I did that. Unless I messed up somewhere, I end up getting [-4, -2]; plugging in -3 does not satisfy the equation, since the absolute value of -1/-2 is not greater than or equal to 1. – Mana Sep 25 '10 at 21:12
@Mana, try to follow your computation backwards to see where the mistake occurred. When you find it, correct it. – Mariano Suárez-Alvarez Sep 25 '10 at 21:24
In particular, if you do not tell us what you did it is impossible for us to help you with what you did! – Mariano Suárez-Alvarez Sep 25 '10 at 21:45
@Mana. When you multiply both sides by $g(x)$ you have to take into account the possibility that it could be negative. Your solution is exactly what you (wrongly) get if you forget this (together with another mistake). – a.r. Sep 25 '10 at 21:50
@Mana. As for your second mistake: notice that Mariano rightly says that the solution is equal to the union of his two inequalities. Plus the fact you were assuming $g(x)$ always to be positive, it seems to me that you were taking the intersection. – a.r. Sep 25 '10 at 22:15

You can also rewrite in equivalent form as

$$f^2(x) \ge g^2(x) \wedge g(x) \neq 0$$


$$ (f(x)+g(x))(f(x)-g(x)) \ge 0 \wedge g(x) \neq 0$$

In your particular case, we get

$$ (3x+6)(x+4) \ge 0 \wedge x+1 \ne 0$$

Which is equivalent to

$$ (x \ge -2 \vee x \le -4) \wedge x \ne -1$$

The $\wedge$ stand for AND and $\vee$ stands for OR, so the solution set is $x$ is not -1 AND either $x \ge -2$ OR $x \le -4$, i.e $$\mathbb{R} - ((-4,-2) \cup \{1\})$$

Since you seem to have the answer reversed, my guess it that you multiplied one of you inequalities by a negative number at some stage and forgot to reverse the inequality.

For instance $$ x < 1$$ Mulitplying by $-1$ gives

$$ -x > -1$$

But if you forget to reverse the inequality it gives $$ -x < -1$$ which is incorrect.

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Not that there is anything wrong with this way of reasoning, but I can't help but wonder why one would use this path. To me it seems as an unnecessary detour and it doesn't really avoid any difficulties. Is there any advantage to this method? – mweerden Sep 25 '10 at 23:33
@mweerden: Well it amounts to finding the points where h(x) = (f(x)+g(x))(f(x)-g(x)) is non-negative. In case f(x) and g(x) are linear, this can easily be done without pen and paper! In some sense, there are fewer cases to consider here. – Aryabhata Sep 25 '10 at 23:40
You can't square both sides of an inequality. It add extraneous roots that cannot always be fully discovered. – Steven Gregory Jun 8 '15 at 20:46
@StevenGregory: You can. Notice the absolute values... – Aryabhata Jun 9 '15 at 0:17

Algebraically, Use the properties of the absolute value function (which properties?) and the difference of two squares to write:

$\frac{|2x+5|}{|x+1|}\geq 1$,

$\Rightarrow |2x+5|\geq |x+1|$, $x\neq -1$.

$\Rightarrow |2x+5|^2\geq |x+1|^2$,

$\Rightarrow (2x+5)^2-(x+1)^2\geq 0$,

$\Rightarrow [(2x+5)+(x+1)][(2x+5)-(x+1)]\geq 0$,

$\Rightarrow (3x+6)(x+4)\geq 0$, $x\neq -1$.

This is a concave up quadratic so positive outside the roots (but not at $-1$). Solution Set is $((-\infty,-4]\cup[-2,\infty))\backslash\{-1\}$.

Geometrically, Our inequality reads:

$\frac{|2x+5|}{|x+1|}\geq 1$.

Now if $x+1\neq0\Leftrightarrow x\neq -1$ (which we have to check separately... yes it can't be a solution), we can multiply both sides by the positive quantity $|x+1|$ to yield $|2x+5|\geq |x+1|$.

Now $2x+5$ is a line of slope $2$ and $y$-intercept $5$, while $x+1$ is a line of slope $1$ and $y$-intercept $1$. Plot these functions[{2x%2B5%2Cx%2B1}], and then take absolutes values by reflecting the negative outputs in the $x$-axis:[{Abs[2x%2B5]%2CAbs[x%2B1]}%2C{x%2C0%2C-6}].

Now it is clear that the solution set is $\mathbb{R}\backslash ((-4,-2)\cup \{-1\})$.

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With the restriction that $x \ne -1$, $\; \left|\dfrac{2x + 5}{x + 1}\right| \geq 1$ is equivalent to

$\dfrac{2x + 5}{x + 1} \ge 1 \qquad$ or $\qquad \dfrac{2x + 5}{x + 1} \le -1$

If $ x + 1 \gt 0$, then

$\quad (2x + 5 \ge x + 1$ and $x > -1)\qquad$ or $\qquad (2x + 5 \le -x - 1$ and $x > -1)$

$\quad (x \ge -4$ and $x > -1)\qquad$ or $\qquad (x \le -2$ and $x > -1)$

$\quad x \in (-1, \infty) \cup \varnothing$

If $ x + 1 \lt 0$, then

$\quad (2x + 5 \le x + 1$ and $x < -1)\qquad$ or $\qquad (2x + 5 \ge -x - 1$ and $x < -1)$

$\quad (x \le -4$ and $x < -1)\qquad$ or $\qquad (x \ge -2$ and $x < -1)$

$\quad x \in (-\infty, -4] \cup [-2, -1)$

$x \in (-\infty, -4] \cup [-2, -1) \cup(-1, \infty)$

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