Rational Inequalities
A component factory wants to know how many pieces it must produce per day so that the average cost per piece drops below a certain price. The average cost is a fraction with the unknown in the denominator. How do you solve an inequality like that? The temptation is to "get rid of" the fraction by cross-multiplying, but that is dangerous: you don't know whether what you are multiplying by is positive or negative. The correct method is to study the sign of the numerator and the denominator with a sign chart.
Why You Don't Cross-Multiply
Solve the following rational inequality, expressing its solution as an interval: $\frac{x-4}{x+3} > 0$
Solve the following rational inequality, expressing its solution as an interval:
$\frac{x-4}{x+3} > 0$
Solve the inequality: $\frac{x}{x-5} < 0$
Solve the inequality:
$\frac{x}{x-5} < 0$
Find the solution set of the inequality: $\frac{2x+6}{x-1} \le 0$
Find the solution set of the inequality:
$\frac{2x+6}{x-1} \le 0$
The Denominator Is Never Zero
Look at the graph of $\frac{x-4}{x+3}$: it is positive where the curve is above the $x$-axis and it does not exist at $x=-3$.
Determine the values of $x$ that satisfy: $\frac{4}{x-6} \le 0$
Determine the values of $x$ that satisfy:
$\frac{4}{x-6} \le 0$
Solve the following inequality: $\frac{x^2-9}{x+1} > 0$
Solve the following inequality:
$\frac{x^2-9}{x+1} > 0$
Solve the rational inequality, paying special attention to multiple roots of the numerator: $\frac{x^2-6x+9}{x+4} \le 0$
Solve the rational inequality, paying special attention to multiple roots of the numerator:
$\frac{x^2-6x+9}{x+4} \le 0$
When There Is a Term on the Other Side
If the inequality is $\frac{P(x)}{Q(x)}\gtrless k$ with $k\neq0$, or has fractions on both sides, first move everything to one side and reduce to a single fraction. The sign chart only compares against zero.
Find the solution set of the inequality: $\frac{x+3}{x-2} \ge 2$
Find the solution set of the inequality:
$\frac{x+3}{x-2} \ge 2$
Determine the set of values that satisfy the following inequality between algebraic fractions: $\frac{1}{x+1} < \frac{1}{x-3}$
Determine the set of values that satisfy the following inequality between algebraic fractions:
$\frac{1}{x+1} < \frac{1}{x-3}$
Find the solution set of the following rational inequality: $\frac{x+2}{x-1} \le \frac{x-4}{x+2}$
Find the solution set of the following rational inequality:
$\frac{x+2}{x-1} \le \frac{x-4}{x+2}$
Factoring, Domains, and Systems
When the numerator and denominator have higher degree, factor them first (common factor, special products, trinomials, synthetic division). That way the roots are in plain view.
For a system of inequalities, solve each one separately and take the intersection of the solutions. Drawing both on parallel number lines helps.
Determine the domain of definition of the function of a real variable $f(x) = \sqrt{\frac{x^2-x-2}{x^2+4x+3}}$.
Determine the domain of definition of the function of a real variable $f(x) = \sqrt{\frac{x^2-x-2}{x^2+4x+3}}$.
Solve the rational inequality, first factorising the polynomials in the numerator and denominator: $\frac{x^4-16}{x^3-2x^2} \le 0$
Solve the rational inequality, first factorising the polynomials in the numerator and denominator:
$\frac{x^4-16}{x^3-2x^2} \le 0$
Solve the following system of rational inequalities and express the result as a union of intervals:…
Solve the following system of rational inequalities and express the result as a union of intervals:
$\begin{cases} \frac{x-2}{x+3} \ge 0 \\ \frac{x^2-16}{x} \le 0 \end{cases}$
Factors That Never Change Sign and "For All x"
Sometimes a factor has no real roots, like $x^2+4$ or $x^2+2x+4$. Then it never changes sign: it adds no points to the chart, and it is enough to know whether it is always positive or always negative.
You can also require that an inequality hold for all $x$. That happens when the numerator has no real roots and has the right sign.
Solve the following rational inequality in which the numerator is a cubic polynomial: $\frac{x^3-8}{x^2-2x-3} \ge 0$
Solve the following rational inequality in which the numerator is a cubic polynomial:
$\frac{x^3-8}{x^2-2x-3} \ge 0$
Determine for which values of the real parameter $k$ the following inequality holds for all $x \in \mathbb{R}$: $\frac{x^2-kx+9}{x^2+1} > 0$
Determine for which values of the real parameter $k$ the following inequality holds for all $x \in \mathbb{R}$:
$\frac{x^2-kx+9}{x^2+1} > 0$
Problems: Average Cost, Speed, Concentration
Rational functions model average cost per unit, time as a function of speed, or the concentration of a drug. The approach is the same: translate the problem statement into a rational inequality and solve it with the sign chart, without forgetting the restrictions of the context.
And the graph confirms it: the average-cost curve lies below the line $y=25$ from $x=80$ on.
The average daily cost (in pesos) of producing $x$ electronic components in a factory is given by the expression $C(x) = \frac{16\,000+300x}{x}$. Determine for which…
The average daily cost (in pesos) of producing $x$ electronic components in a factory is given by the expression $C(x) = \frac{16\,000+300x}{x}$. Determine for which production levels the average cost will be strictly less than $500$ pesos per component.
The time $T$ (in hours) it takes a boat to travel a distance of $120$ km downstream is modelled by $T(v) = \frac{120}{v+4}$, where $v$ is the boat's own speed in km/h.…
The time $T$ (in hours) it takes a boat to travel a distance of $120$ km downstream is modelled by $T(v) = \frac{120}{v+4}$, where $v$ is the boat's own speed in km/h. For which speeds $v > 0$ will the journey take less than $5$ hours?
The concentration of a drug in the bloodstream, measured in mg/L, $t$ hours after its administration is given by the function $C(t) = \frac{4t}{t^2+4}$. During which…
The concentration of a drug in the bloodstream, measured in mg/L, $t$ hours after its administration is given by the function $C(t) = \frac{4t}{t^2+4}$. During which time interval $t \ge 0$ is the concentration of the drug greater than or equal to $1$ mg/L?
In optics, the thin lens equation relates the focal length $f$, the object distance $p$ and the image distance $q$ by the formula…
In optics, the thin lens equation relates the focal length $f$, the object distance $p$ and the image distance $q$ by the formula $\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$. If a lens with a focal length of $f = 12$ cm is used, determine for which object distances $p > 0$ the image distance $q$ will be greater than $24$ cm.
Back to the beginning: the average cost drops below a certain price exactly when the corresponding rational inequality holds, and that is decided by the sign of the numerator and the denominator, not by cross-multiplying.