Solving inequalities with "x" in the denominator has always been a stumbling block for me. Other than understanding how a particular expression, such as 1/x, works (in this case, x cannot be zero), how might I go about solving inequalities having such expressions mathematically.
For instance:
1/x < 0
x(1/x) < x(0)
1 < 0 // This is where things seem to break down. // Probably because of the vertical asymptote.As a slightly longer example:
1/x < 4
1/x - 4 < 0
x(1/x - 4) < x(0)
1 - 4x < 0
-4x < -1
x > 1/4 // Makes sense up to this point (Only covers x > 0) // How might I now solve for a negative x value?I attempted changing the LessThan operator to a GreaterThan operator as per the rule:
if a < b and c < 0, then ac > bcyet this resulted in an incorrect answer.
I already know that the answer to the inequality is (-infinity, 0) union (1/4, infinity). I would just like to know how to solve this algebraically.
$\endgroup$6 Answers
$\begingroup$Solving for when $\frac{1}{x}<4$.
First, we do not allow $x=0$. Let's split into cases.
Case 1: $x>0$. Then $\frac{1}{x}<4$ becomes $1< 4x$ or $\frac{1}{4}<x$. This means that whenever both $x>0$ and $x>\frac{1}{4}$ the inequality holds. Combining these inequalities, we get that it holds for $x>\frac{1}{4}$. (I know it does nothing here, but sometimes this is important)
Case 2: $x<0$. Then $\frac{1}{x}<4$ becomes $1> 4x$ or $\frac{1}{4}>x$. This means that whenever both $x<0$ and $x<\frac{1}{4}$ the inequality holds. Combining these inequalities, we get that it holds for $x<0$.
Now, combining both cases we get that it holds for $x<0$ or $x>\frac{1}{4}$.
$\endgroup$ 1 $\begingroup$Here's one way. $1/x\lt4$, multiply both sides by $x^2$, $x\lt4x^2$, $4x^2-x\gt0$, $x(4x-1)\gt0$, so either $x\gt0$ and $4x-1\gt0$, or $x\lt0$ and $4x-1\lt0$, etc.
$\endgroup$ 1 $\begingroup$Breaking into cases is unnecessary and unwieldy. Consider the inequality: $x<\frac{1}{x}$. We subtract $x$ from both sides: $0< \frac{1}{x} -x$. Now we find a common denominator on the right:$ 0 < \frac{1-x^2}{x}$. Factor the numerator: $0<\frac{(1-x)(1+x)}{x}.$ The expression may change signs at $x=-1,0,$ or at $x=1$. The numerator represents a parabola that spills water. The totality of signs goes like this:
$$\begin{array}{ccccccc} + & u & - & 0 & + & u& - \\ x<-1 & x=-1 & -1<x<0 & 0 & 0<x<1 & x=1 & 1<x \end{array}$$ In general, you make the expression comparable to $0$, factor numerator and denominator, and examine how the expression changes at the intercepts (numerator =0) and asymptotes (denominator =0) See for more details. Specifically, here .
$\endgroup$ 3 $\begingroup$It is generally a good idea to multiply both sides by a positive number. In this case solving $\frac{1}{x} \gt 4$, we multiply by $x^2$ which is always positive. thus we get $$4x^2-x \lt 0$$ upon rearranging. the solution is thus $$0\lt x \lt \frac{1}{4}.$$
$\endgroup$ 2 $\begingroup$Concerning the first example: If $1/x < 0$, then $x < 0$. Now, multiplying by a negative number changes the direction of the inequality sign, and hence $x(1/x) > x (0)$ ($1 > 0$).
$\endgroup$ $\begingroup$See my answer to this question for a description of a general method to solve inequalities between rational functions. It also describes the relationship between multiplying through by a least common denominator vs. the square of such.
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