Pre-Calc Diagnostic: Solve Each Inequality in Interval Notation (Stewart Problem 9)

Direct answer. Stewart's Problem 9 is the inequality companion to Problem 8 — five inequalities, five shapes, five first moves, and every answer written in interval notation: (a) −4 < 5 − 3x ≤ 17 is a linear compound inequality with a sign flip when dividing by −3 ([−4, 3)); (b) x² < 2x + 8 is a quadratic, rewrite as x² − 2x − 8 < 0, factor, and test ((−2, 4)); (c) x(x − 1)(x + 2) > 0 is a higher-degree product, draw a sign chart ((−2, 0) ∪ (1, ∞)); (d) |x − 4| < 3 is an absolute value, unpack as a compound inequality ((1, 7)); and (e) 2x − 3x + 1 ≤ 1 is a rational inequality, bring to one side and find critical points ((−1, 4]). Every answer is written with brackets/parentheses chosen by whether each endpoint is included.

The interval notation cheat sheet — the part every student should have automatic

Before any algebra, every pre-calc student needs four conventions memorized:

• Square bracket [ or ] means the endpoint is included — corresponds to ≤ or ≥.
• Round parenthesis ( or ) means the endpoint is excluded — corresponds to < or >.
• Infinity ∞ always gets a parenthesis. You cannot include infinity.
• Disjoint intervals are joined with the union symbol ∪. The smaller number always goes on the left.

Mixing up ( and [ at an endpoint is the most common error on this entire problem — and it almost always loses the full point on AP Calc free-response when the question asks for an interval of increase or a domain. Get the brackets right.

(a) Linear compound inequality with a sign flip: −4 < 5 − 3x ≤ 17

A compound inequality is two inequalities chained together. The strategy is to isolate x in the middle by applying the same operation to all three parts. Start by subtracting 5 everywhere:

−4 − 5 < −3x ≤ 17 − 5

−9 < −3x ≤ 12

Now divide all three parts by −3. Because we divided by a negative, both inequality signs flip:

3 > x ≥ −4

Rewrite in the usual left-to-right small-to-large order:

−4 ≤ x < 3

[−4, 3)

Square bracket on −4 because the original was ≤ after flipping (an = case is allowed). Parenthesis on 3 because the original was strict < after flipping (5 − 3x can never quite equal −4 under a strict inequality, which corresponds to x never quite equaling 3). The sign-flip rule is the single most-cited reason for lost points on the inequality section. Drill it.

(b) Quadratic by factoring and sign analysis: x² < 2x + 8

Never solve a quadratic inequality by "doing the same thing to both sides." Always move everything to one side so you have an expression compared to zero:

x² − 2x − 8 < 0

Factor. Two integers that multiply to −8 and add to −2: −4 and +2.

(x − 4)(x + 2) < 0

The product of two factors is negative exactly when the two factors have opposite signs. The roots −2 and 4 split the number line into three regions; the product is positive on the outside intervals and negative on the middle one.

(x − 4):   − − − − −   0   + + + + +
(x + 2):   − − −   0   + + + + + + + + +
product:     + + +   0   − − −   0   + + + + +
                            −2           4

We want the product less than zero — the middle interval. The endpoints are not included because the inequality is strict.

(−2, 4)

Pictorially: the parabola y = (x − 4)(x + 2) opens upward and dips below the x-axis between its two roots. That dip is the answer.

(c) Higher-degree product with a sign chart: x(x − 1)(x + 2) > 0

Three factors, three roots: x = −2, 0, 1. The number line splits into four intervals. The sign of each factor flips at its own root, so the product's sign alternates across the intervals.

(x + 2):   − −   0   + + + + + + + + + + + + + + +
    x:   − − − − − − − −   0   + + + + + + + +
(x − 1):   − − − − − − − − − − − − −   0   + + +
product:   − −   0   + + + + +   0   − − − −   0   + + +
                      −2           0         1

We want the product strictly greater than zero, so we take the two intervals where the product is positive. Endpoints are excluded because of strict >.

(−2, 0) ∪ (1, ∞)

The shortcut once you've drawn enough of these: count the negatives in each interval. An even number of negative factors gives a positive product; odd gives negative. From left to right starting at −∞, all three factors are negative (three negatives → negative); cross −2, now two negatives → positive; cross 0, one negative → negative; cross 1, zero negatives → positive. Same answer, faster.

(d) Absolute value as distance: |x − 4| < 3

Read the inequality as a distance statement: |x − 4| is the distance from x to 4 on the number line, and we want that distance to be less than 3. So x lies within 3 units of 4 — between 1 and 7. Formally, |A| < B (with B > 0) unpacks as the compound inequality −B < A < B:

−3 < x − 4 < 3

Add 4 to all three parts:

1 < x < 7

(1, 7)

Both endpoints excluded because of strict <. The mirror rule is worth memorizing alongside this one: |A| > B splits into A < −B or A > B, which gives a union of two rays heading outward to infinity — not a single bounded interval. Less-than absolute value gives a bounded interval; greater-than gives the complement.

(e) Rational inequality with a critical-point and domain check: 2x − 3x + 1 ≤ 1

The single most-trapped move on this entire problem: students multiply both sides by x + 1 to clear the denominator. Do not do this. The sign of x + 1 depends on x — sometimes positive, sometimes negative — so the inequality may or may not flip, and you cannot tell which without already knowing where x lives. Instead, bring everything to one side and combine over a common denominator:

2x − 3x + 1 − 1 ≤ 0

2x − 3 − (x + 1)x + 1 ≤ 0   ⟹   x − 4x + 1 ≤ 0

Now we have a single rational expression compared to zero. Identify the critical points: the zero of the numerator gives x = 4, and the zero of the denominator gives x = −1 (and at that point the expression is undefined, so it can never be included in the answer). These two critical points split the number line into three intervals.

(x − 4):   − − − − − − − − − − − − −   0   + + +
(x + 1):   − −   0   + + + + + + + + + + + + + + +
ratio:       + +  und   − − − − − − −   0   + + + +
                    −1                   4

We want the ratio less than or equal to zero. The middle interval gives the ratio negative, and at x = 4 the ratio equals zero — which is allowed because of ≤. At x = −1 the ratio is undefined, so that endpoint is excluded no matter what:

(−1, 4]

Parenthesis on −1 because the expression is undefined there. Square bracket on 4 because the numerator is zero (so the whole fraction equals zero) and the inequality allows equality. This bracket-choice logic is exactly what AP Calc graders look for when they ask for "the interval where f'(x) ≤ 0" — open vs. closed endpoints are graded individually.

The one big-picture lesson: every inequality answer is a sign analysis

Every problem on Stewart's diagnostic Problem 9 reduces to the same underlying question: where is some expression positive, negative, or zero? Once a pre-calc student internalizes that and gets fluent at building a quick sign chart, every inequality on the test — linear, polynomial, rational, absolute value — becomes routine. The five shapes worth memorizing:

• Linear inequality (or compound): isolate x like an equation, but flip every sign when you multiply or divide by a negative.
• Polynomial inequality: move everything to one side, factor, find the roots, build a sign chart, take the intervals matching the inequality direction.
• Higher-degree product: same as above — count negative factors in each region as a sign-chart shortcut.
• Absolute value: |A| < B → compound inequality (bounded). |A| > B → union of two rays (unbounded).
• Rational inequality: NEVER multiply both sides by the denominator. Bring to one side, combine over a common denominator, identify zeros and undefined points as critical points, build a sign chart, choose intervals.

These five patterns cover everything AP Calc AB and BC test on inequality reasoning — domains, intervals of increase and decrease, concavity intervals, where a function is positive or negative. A pre-calc student who has them automatic walks into Calc with a real advantage.

How SOMATH teaches the inequality toolkit on the Upper West Side

We start every new pre-calc student with a diagnostic identical in structure to Stewart's. Five inequalities of different shapes, no calculator, no help. Then we sit with the student and ask — before they pick up a pencil — what shape each one is, what the first move is, and how the answer should be written. The bracket discipline is graded as strictly as the algebra. From our classroom on 226 W 79th St we run small groups of 3–5 students, every lesson on whiteboards, every sign chart drawn out loud. Every new student gets a free 30-minute evaluation and a written diagnostic within 48 hours, even if they don't enroll.

FAQ

Why flip the sign when dividing by a negative?

Because multiplying or dividing by a negative reverses the order of every pair of real numbers. a < b implies −a > −b.

How do I write an answer in interval notation?

Square bracket if the endpoint is included (≤ or ≥), parenthesis if excluded (<, >, or infinity). Disjoint intervals joined with ∪.

When do I need a sign chart?

Anytime you have a polynomial inequality with more than one root, or any rational inequality. The sign chart determines which intervals satisfy the inequality.

How is a rational inequality different from a rational equation?

You cannot multiply by the denominator on a rational inequality, because the denominator's sign may change. Bring to one side, combine over a common denominator, then sign-chart.

Why does |x − 4| < 3 unpack into a compound inequality?

Because |x − 4| is the distance from x to 4, and "distance less than 3" means x is between 1 and 7.

Related posts

• Pre-Calc Diagnostic: Solve the Equation — 7 Techniques (Stewart Problem 8)
• Pre-Calc Diagnostic: Rewrite by Completing the Square (Stewart Problem 7)
• Pre-Calc Diagnostic: Rationalize the Expression (Stewart Problem 6)