D9 - Rational inequalities

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Multiplying or dividing an inequality by a positive number preserves its direction. Multiplying or dividing by a negative number reverses it. A rational expression is defined only where its original denominator is nonzero.

Worked example 1: comparing with a constant

Question. Write $\frac{x+1}{x-2}>2$ as an equivalent single-fraction inequality with zero on the right. State its domain.

Plan. Subtract 2 and express it with denominator $x-2$, keeping the restriction $x\ne2$.

Answer. $\frac{5-x}{x-2}>0$, with $x\in\mathbb R$, $x\ne2$.

Check. At $x=3$, the original left side minus 2 is $4-2=2$. The new fraction gives $(5-3)/(3-2)=2$. At $x=0$, both differences are $-5/2$, so the negative-denominator case is retained too.

Worked example 2: comparing with a line

Question. Write $\frac{x+1}{x-2}>x$ as an equivalent single-fraction inequality with zero on the right. State its domain.

Plan. Subtract $x$, then put both terms over $x-2$.

No multiplication of the inequality by $x-2$ has occurred: this is an identity between differences on the original domain.

Answer. $\frac{-x^2+3x+1}{x-2}>0$, with $x\ne2$.

Check. At $x=-1$, the original difference is $0-(-1)=1$. The new fraction is $(-1-3+1)/(-3)=1$. This also checks the sign in front of $x(x-2)$.

Watch out

Wrong: $x+1-x(x-2)=x+1-x^2-2x$. The minus sign acts on the entire product. Correct: $x+1-x^2+2x=-x^2+3x+1$.

Critical values of a rational expression

For a rational inequality written as one fraction compared with zero, its critical values are the real numerator zeros and original denominator zeros. A numerator zero may be included for a non-strict inequality; an original denominator zero is never included.

Worked example 3: one zero and one exclusion

Question. For $r(x)=\frac{x-2}{x+1}$, find its zero and excluded input. Describe the sign behaviour near each.

Plan. Solve numerator and denominator equations separately, then inspect the signs immediately on either side.

At 2, the numerator changes sign while the denominator stays positive. At $-1$, the denominator changes sign while the numerator stays negative. The fraction changes sign across both critical values, but it equals zero only at 2.

Answer. Zero at $x=2$; domain $\mathbb R\setminus\{-1\}$. Both critical values separate opposite signs.

Check. $r(2)=0/3=0$. The nearby inputs $-2,0$ give 4 and $-2$, whereas inputs 1 and 3 give $-1/2$ and $1/4$. These verify the two sign reversals without evaluating at the excluded input.

Worked example 4: an even numerator zero

Question. For $s(x)=\frac{(x-2)^2}{x+1}$, find its zero and excluded input. Describe the sign behaviour near each.

Plan. Keep the original denominator restriction and use the positive square away from 2.

Away from 2, the fraction therefore has the sign of $x+1$. It changes sign at $-1$, but is positive on both sides of 2.

Answer. Zero at $x=2$; domain $\mathbb R\setminus\{-1\}$. There is a sign reversal at $-1$, but none at 2.

Check. $s(1)=1/2$ and $s(3)=1/4$, both positive. Also $s(-2)=-16$ while $s(0)=4$, confirming the reversal across $-1$.

Key idea

“Not negative” allows zero; it does not allow an undefined value. Equality in an inequality never restores an excluded denominator input.

Worked example 5: a non-strict rational inequality

Question. Solve $\frac{x-2}{x+1}\ge0$.

Plan. Split at $-1$ and 2, compare signs, then decide which boundaries belong.

The positive intervals are $x<-1$ and $x>2$. Include 2 because it gives zero; exclude $-1$ because the fraction is undefined there.

Answer. $( -\infty,-1)\cup[2,\infty)$.

Check. $r(x)=1-\frac3{x+1}$. At $x=-2,0,3$, this gives 4, $-2$, $1/4$, independently checking all three interval signs.

Worked example 6: keeping a zero inside an allowed interval

Question. Solve $\frac{(x-2)^2}{x+1}\ge0$.

Plan. Use the squared numerator's sign, but check its zero separately.

The table covers open intervals. At 2 the fraction is zero, so it belongs and joins the two positive intervals. The input $-1$ remains excluded.

Answer. $x>-1$, or $(-1,\infty)$.

Check. For $x>-1$, the denominator is positive and the numerator is non-negative, proving that the whole interval works. For $x<-1$, the numerator is positive and the denominator negative, so none works.

Watch out

Wrong: signs must alternate at every critical value. An even power does not change sign. Inspect multiplicities or use test values rather than forcing an alternating pattern.

Worked example 7: comparing a fraction with a line

Question. Solve $\frac{x+1}{x-2}>x-1$.

Plan. Form one fraction, multiply by the positive square $(x-2)^2$, and use exact polynomial signs.

The numerator is $3-(x-2)^2$, so its zeros are $2-\sqrt3$ and $2+\sqrt3$. Its sign is positive between these and negative outside. Combining with $x-2$ gives product signs $+,-,+,-$ across the ordered boundaries $2-\sqrt3,2,2+\sqrt3$.

Answer. $x<2-\sqrt3$ or $2<x<2+\sqrt3$.

Check. At $x=0$, the original comparison is $-1/2>-1$, true. At $x=1$, it is $-2>0$, false. At $x=3$, it is $4>2$, true; at $x=4$, $5/2>3$, false. These test the four intervals directly.

Worked example 8: the other direction

Question. Solve $\frac{x+1}{x-2}<x-1$.

Plan. Use the same difference, keeping the sign reversal in the question distinct from any multiplication step.

The four signs are still $+,-,+,-$; select the second and fourth intervals.

Answer. $2-\sqrt3<x<2$ or $x>2+\sqrt3$.

Check. Split directly at 2. For $x<2$, multiplication by $x-2<0$ gives $x+1>(x-1)(x-2)$, hence $(x-2)^2<3$; intersection with $x<2$ gives $2-\sqrt3<x<2$. For $x>2$, the direction stays $<$, giving $(x-2)^2>3$, hence $x>2+\sqrt3$. The case method agrees.

Watch out

Wrong: $N(x)/D(x)\le0$ means $N(x)D(x)\le0$ for every real $x$. The equivalence holds only for $D(x)\ne0$. The new polynomial vanishes at denominator zeros, but the original fraction is undefined there.