Solving inequalities

An inequality asks for a set of inputs. Factor a rational expression, mark every numerator zero and denominator zero, and test the intervals between them. For $b>0$, $|x-a|<b$ means $a-b<x<a+b$, while $|x-a|>b$ selects the two outside intervals.

Critical values and repeated roots

Factor the polynomial, mark every real zero and test signs between them. For $x^2+2x-15=(x+5)(x-3)$, the roots are $-5,3$ and interval signs are $+,-,+$. Thus $x^2+2x-15<0$ gives $-5<x<3$, where the upward-opening parabola is below the axis.

For $(x-1)(x+1)(x+3)^2$, signs on $(-\infty,-3),(-3,-1),(-1,1),(1,\infty)$ are $+,+,-,+$. The double root $-3$ does not change the sign. The positive inequality gives $x<-3$ or $-3<x<-1$ or $x>1$; the nonpositive inequality gives $x=-3$ or $-1\le x\le1$.

Example 1’s test-point paragraph prints $(x-5)(x+3)$: that is a typo. Test the original $(x+5)(x-3)$, consistent with the source roots and final answer.

Nine rules for preserving an inequality

Before cancelling a factor, check its sign. When applying a function to both sides, check whether it increases or decreases.

1. Transitivity: $a<b$ and $b<c$ imply $a<c$. For example, $-1<2<13$ gives $-1<13$.
2. Add the same number: $a<b$ implies $a+c<b+c$. Starting with $2<7$, adding $4$ gives $6<11$; adding $-4$ gives $-2<3$.
3. Multiply by a positive number: if $a>b$ and $c>0$, then $ac>bc$. From $2>-1$, multiplication by $3$ gives $6>-3$.
4. Multiply by a negative number: if $a>b$ and $c<0$, then $ac<bc$. Multiplying $2>-1$ by $-3$ gives $-6<3$. Check this sign before multiplying by an expression involving $x$.
5. Square positive inputs: $a>b>0$ implies $a^2>b^2$. For example, $3>2$ gives $9>4$. Do not drop the positivity condition.
6. A positive product: $ab>0$ means both factors are positive or both are negative. For example, $(2)(3)>0$ and $(-2)(-3)>0$.
7. A negative product: $ab<0$ means the factors have opposite signs. For example, $(2)(-3)<0$ and $(-2)(3)<0$.
8. Apply an increasing function: $a<b$ implies $f(a)<f(b)$ when $f$ is strictly increasing. The examples are $\ln x$ on positive inputs and $e^x$: $\ln a<\ln b$ and $e^a<e^b$.
9. Apply a decreasing function: $a<b$ implies $f(a)>f(b)$ when $f$ is strictly decreasing. For $f(x)=e^{-x}$, this gives $e^{-a}>e^{-b}$.

Algebraic and graphical methods describe the same solution set. Check critical values individually for non-strict inequalities; an even repeated root need not change the sign. When asked for the set of values, present set notation. A range may be written with inequalities, interval notation or set notation.

Choose one of three methods

One graph: make the right side zero, for example $f(x)-g(x)>0$. Sketch the difference; find its x-intercepts and vertical asymptotes, then read the required intervals.

Two graphs: sketch $y=f(x)$ and $y=g(x)$. Find intersection inputs and vertical asymptotes of either graph, then compare which graph lies higher.

Analytical: make the right side zero, combine the left side into one rational expression and factor numerator and denominator as far as possible. Complete the square when a quadratic does not factor over the reals. Mark critical values and test every interval.

For other expressions a graph is often useful. Always inspect domain endpoints and vertical asymptotes as well as crossings. When squaring a radical comparison, first check that its radicand is nonnegative and that the two quantities being compared have the required signs.

Modulus: a nonnegative magnitude

By definition, $|x|=x$ for $x\ge0$ and $|x|=-x$ for $x<0$. Its graph reflects the negative half of $y=x$ above the axis.

• $|x|\ge0$ for every real $x$.
• $|x|^2=|x^2|=x^2$.
• $\sqrt{x^2}=|x|$, which need not equal $x$ when $x<0$.
• For a positive constant $a$, $|x|<a$ means $-a<x<a$.
• For a positive constant $a$, $|x|>a$ means $x<-a$ or $x>a$.

Compare the V graph with $y=a$ to see the inside and outside intervals. Simple modulus inequalities can use these properties directly; more complicated ones can use graph comparisons. The positivity condition matters: $|x|<-2$ has no solution, whereas $|x|>-2$ holds for every real $x$.

Recognise a hence substitution

Match the complete expression to a form already solved. Every occurrence of the old input must be replaced, including powers, denominators and modulus bars. Common replacements in these notes are $x^2$, $e^x$, $\ln x$, $|x|$ and $\sin x$.

Then intersect the resulting condition with the replacement’s range and domain: $|x|\ge0$, $e^x>0$, $\ln x$ requires $x>0$, and $-1\le\sin x\le1$. Keep denominator exclusions after substitution. Examples 10 and 11 show why the algebraic answer alone is not the last step.