10.4 Rational Inequalities
When dealing with a rational inequality, never cross-multiply.
The reason is the rule at the head of the chapter. Cross-multiplying \(\frac {x}{x+2}\geq 3\) means multiplying both sides by \(x+2\), whose sign is not known — it is positive when \(x>-2\) and negative when \(x<-2\) — so there is no way to know whether the inequality should keep its direction or reverse it. Doing it anyway gives \(x\geq 3x+6\), hence \(x\leq -3\), which is wrong: the true answer is \([-3,-2)\), and the value \(x=-4\) that this method admits does not satisfy the original, since \(\frac {-4}{-2}=2\) is not \(\geq 3\).
The reliable procedure is always the same:
- (i).
- move everything to one side, leaving a single fraction compared with \(0\);
- (ii).
- factorise numerator and denominator and find the critical values — the points where either vanishes;
- (iii).
- test the sign of the whole expression on each interval between them.
Points where the denominator vanishes are always excluded, even when the inequality is \(\leq \) or \(\geq \), because
the expression is undefined there. That is why the answers below have a square bracket at one end and
a round one at the other.
Solution.
The best method to solve such inequality is by use of table \begin {align*} \frac {x}{x+2} &\geq 3\implies \frac {x}{x+2}-3\geq 0\\\\ \implies \frac {x-3(x+2)}{(x+2)} & \geq 0\\ \implies \frac {x-3x-6}{x+2} &\geq 0\\ \implies \frac {-2x-6}{x+2} &\geq 0\\ \implies -\left (\frac {2x+6}{x+2}\right ) & \geq 0\\\\ \implies \frac {2x+6}{x+2}\leq 0\\ \end {align*}
Note the factors are \(2x+6\) and \(x+2\).
Therefore the critical values are
\(2x+6=0\implies x=-3\) \(x+2=0\implies x=-2\)
| \((-4)\) | \(-3\) | \((\frac {-5}{2})\) | \(-2\) | 0 | |
| \(2x+6\) | \(-ve\) | \(|\) | \(+ve\) | \(|\) | \(+ve\) |
| \(x+2\) | \(-ve\) | \(|\) | \(-ve\) | \(|\) | \(+ve\) |
| \(\frac {2x+6}{x+2}\) | \(+ve\) | \(|\) | \(-ve\) | \(|\) | \(+ve\) |
\[SS:[-3,-2)\]
Solve the following inequalities
- (a).
- \(\frac {x+1}{x}\leq \frac {1}{3}\)
- (b).
- \(\frac {2x+1}{x-3}<\frac {1}{2}\)
Solution.
\begin {align*} \frac {x+1}{x} & \leq \frac {1}{3}\\ \implies \frac {x+1}{x}-\frac {1}{3} & \leq 0\\ \implies \frac {3(x+1)-x}{3x} & \leq 0\\ \implies \frac {3x+3-x}{3x} & \leq 0\\ \implies \frac {2x+3}{3x} & \leq 0\\ \end {align*}
The critical values, \(2x+3=0\implies x=\frac {-3}{2}\) and \(3x=0\implies x=0\).
| \((-2)\) | \(-\frac {3}{2}\) | \((-1)\) | 0 | \((1)\) | |
| \(2x+3\) | \(-ve\) | \(|\) | \(+ve\) | \(|\) | \(+ve\) |
| \(3x\) | \(-ve\) | \(|\) | \(-ve\) | \(|\) | \(+ve\) |
| \(\frac {2x+3}{3x}\) | \(+ve\) | \(|\) | \(-ve\) | \(|\) | \(+ve\) |
\[SS:\left [-\frac {3}{2},0\right )\]
\begin {align*} \frac {2x+1}{x-3} & < \frac {1}{2}\\ \implies \frac {2x+1}{x-3}-\frac {1}{2} & <0\\ \implies \frac {2(2x+1)-(x-3)}{2(x-3)} & <0\\ \implies \frac {4x+2-x+3}{2(x-3)} & <0\\ \implies \frac {3x+5}{2(x-3)} & <0 \end {align*}
Critical values
\(3x+5=0\implies x=-\frac {5}{3}\) and \(2(x-3)=0\implies x=3\).
| \((-2)\) | \(-\frac {5}{3}\) | 0 | 3 | \((4)\) | |
| \(3x+5\) | \(-ve\) | \(|\) | \(+ve\) | \(|\) | \(+ve\) |
| \(2(x-3)\) | \(-ve\) | \(|\) | \(-ve\) | \(|\) | \(+ve\) |
| \(\frac {3x+5}{2(x-3)}\) | \(+ve\) | \(|\) | \(-ve\) | \(|\) | \(+ve\) |
\[SS:\left (-\frac {5}{3},3\right )\]
For \(\frac {2x+1}{x-3}>\frac {1}{2}\), the solution set is \(SS:\left (-\infty ,-\frac {5}{3}\right )\cup (3,\infty )\).
Find the solution set of the inequality \(\bigg |\frac {2x}{x-3}\bigg |<\frac {1}{2}\).
Solution.
Implies that \(-\frac {1}{2}<\frac {2x}{x-3}<\frac {1}{2}\)
This gives two inequalities to be solved \(-\frac {1}{2}<\frac {2x}{x-3}\) and \(\frac {2x}{x-3}<\frac {1}{2}\).
For \(-\frac {1}{2}<\frac {2x}{x-3}\) we have \(0<\frac {2x}{x-3}+\frac {1}{2}\)
\(\implies 0<\frac {4x+x-3}{2(x-3)}\implies 0<\frac {5x-3}{2(x-3)}\)
The critical values are \(x=\frac {3}{5}\) and \(x=3\).
| 0 | \(\frac {3}{5}\) | \((2)\) | 3 | \((4)\) | |
| \(5x-3\) | \(-\) | \(|\) | \(+\) | \(|\) | \(+\) |
| \(2(x-3)\) | \(-\) | \(|\) | \(-\) | \(|\) | \(+\) |
| \(\frac {5x-3}{2x-6}\) | \(+\) | \(|\) | \(-\) | \(|\) | \(+\) |
\[SS_1:\left (-\infty ,\frac {3}{5}\right )\cup (3,\infty )\]
For \(\frac {2x}{x-3}<\frac {1}{2}\), we have that \(\frac {2x}{x-3}-\frac {1}{2}<0\)
\[\implies \frac {4x-x+3}{2(x-3)}<0\implies \frac {3x+3}{2(x-3)}<0\]
critical values: \(x=-1\) and \(x=3\).
| \((-2)\) | \(-1\) | 0 | 3 | \((4)\) | |
| \(3x+3\) | \(-\) | \(|\) | \(+\) | \(|\) | \(+\) |
| \(2x-6\) | \(-\) | \(|\) | \(-\) | \(|\) | \(+\) |
| \(\frac {3x+3}{2x-6}\) | \(+\) | \(|\) | \(-\) | \(|\) | \(+\) |
\[SS_2:(-1,3)\]
The required solution is the intersection of \(SS_1\) and \(SS_2\).
\[SS:\left (-1,\frac {3}{5}\right )\]
Solution.
\[\frac {x}{x^2-4}<0\implies \frac {x}{(x-2)(x+2)}<0\] Critical values: \(x=0, x=-2, x=2\)
| \((-3)\) | \(-2\) | \((-1)\) | 0 | \((1)\) | 2 | \((3)\) | |
| \(x\) | \(-\) | \(|\) | \(-\) | \(|\) | \(+\) | \(|\) | \(+\) |
| \(x-2\) | \(-\) | \(|\) | \(-\) | \(|\) | \(-\) | \(|\) | \(+\) |
| \(x+2\) | \(-\) | \(|\) | \(+\) | \(|\) | \(+\) | \(|\) | \(+\) |
| \((x-2)(x+2)\) | \(+\) | \(|\) | \(-\) | \(|\) | \(-\) | \(|\) | \(+\) |
| \(\frac {x}{(x-2)(x+2)}\) | \(-\) | \(|\) | \(+\) | \(|\) | \(-\) | \(|\) | \(+\) |
\[SS:(-\infty ,-2)\cup (0,2)\]
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