Math 30-1 common mistakes
Math 30-1 · Understand the traps and how to avoid them.
Asked for the inverse of f(x) = (x − 3)² + 1 on x ≥ 3, writes y = ±√(x − 1) + 3 and declares both branches (or just one, arbitrarily) to be the inverse function, omitting the stated domain and range of the inverse.
Inverses are drilled as a three-step procedure - swap, solve, relabel - with the one-to-one requirement treated as an introductory remark rather than a check to perform. Because the algebra 'works', the ± branch and the lost restriction never feel like an error.
What to do instead
Before swapping, run a horizontal-line test and write the restricted domain of f explicitly. After solving, select the branch whose range equals the original domain, then state domain and range of the inverse. Practise on quadratics (complete the square first), absolute value, and y = x² − 4x + 1 to build the habit.
Graphs y = f(2x − 6) by shifting right 6 and then stretching horizontally by 1/2, reading the horizontal parameters as though they were independent of the coefficient on x.
The bracket (2x − 6) looks like the familiar vertex form (x − h), so the (x − h) reading is reused even when a coefficient b is present. The order of operations inside the argument is invisible in the notation, and sketching software hides the mistake.
What to do instead
Force the habit of factoring b out of the whole argument first: f(2(x − 3)), so the shift is 3, not 6. Then confirm with a mapping rule - take three key points and apply (x, y) → (x/b + h, ay + k) - and check that the transformed points lie on the drawn graph.
Solves a radical or rational equation, reports every root the algebra produces, and never states restrictions (denominator ≠ 0, radicand ≥ 0) or discards the extraneous roots introduced by squaring.
Squaring both sides is the natural first move and yields a polynomial equation that looks perfectly valid. Restriction-checking is a separate step that earns no visible marks until it does, so it is the first thing dropped under time pressure.
What to do instead
Write the restrictions as the first line of the solution, before any algebra. After solving, substitute every candidate into the original equation - not the squared one - and record rejected values with the reason. Use one worked example where squaring genuinely produces a false root to make the need concrete.
Solves 2 sin x cos x = sin x by dividing both sides by sin x, losing the entire family sin x = 0, or reports only the principal-value solution when the domain is 0 ≤ x < 2π.
Dividing by a common factor reduces the equation to a familiar linear form and feels like simplification. The general solution and reference-angle symmetry are not rehearsed as a routine, so the second and third solutions are never generated.
What to do instead
Never divide by an expression that can equal zero - factor and set each factor to zero instead. Then use the reference angle with the general forms x = θ + 2πn and x = π − θ + 2πn (or a CAST/quadrant sketch), listing every solution inside the stated interval before writing the final answer.
Reads the period of y = 4 sin(3x − π) + 2 as 6π, and states the phase shift as π rather than π/3, because the 3 has not been factored out of the argument.
The parameters c and b compete inside the bracket. Students memorise 'period = 2π/b' as a formula but then apply the phase shift as if it were a simple (x − h), so one parameter is handled correctly and the other is not.
What to do instead
Rewrite as 4 sin(3(x − π/3)) + 2 before reading anything: amplitude 4, period 2π/3, phase shift π/3 right, vertical displacement 2. Verify by tracing one full cycle from maximum to next maximum on the sketch and checking it spans one period.
Proves a trigonometric identity by cross-multiplying, moving terms from one side to the other, or manipulating both sides simultaneously and then writing LS = RS, having effectively assumed the result.
These are exactly the moves that solve equations, and equation-solving has had far more practice. The 'work one side only' convention reads as a stylistic preference rather than a logical requirement, so it is abandoned when the proof gets hard.
What to do instead
Fix a starting side - usually the more complicated one - and transform it alone until it equals the other. Keep a reference card of reciprocal, quotient, and Pythagorean identities, and when stuck convert everything to sines and cosines or factor to create a Pythagorean pair.
Expands log(x + 5) as log x + log 5, rewrites (log x)² as 2 log x, or cancels log from both sides of an equation without first confirming the arguments are positive.
The product, quotient, and power laws look like ordinary algebraic distribution, and the notation log(x + 5) resembles a coefficient-times-variable pattern. Students learn the three laws without the conditions attached to them.
What to do instead
Disprove each misuse numerically: log(2 + 8) = 1 while log 2 + log 8 ≈ 0.903, so the sum law fails. Write the three valid laws together with their conditions (a, b > 0; base > 0, ≠ 1) on one card, and state the domain x > 0 as the first line of every log equation.
Solves log(x + 3) + log(x) = 1 and accepts both x = 2 and x = −5, or takes the logarithm of only one side when solving an exponential equation such as 3^(2x) = 12.
The domain restriction on logarithms is stated once and never re-tested against answers. Applying an operation to one side only is a common slip when the other side already contains a log or a clean constant.
What to do instead
For a^x = b, take log of both sides and bring the exponent down: x log a = log b. For any log equation, substitute each root back into the original and reject any value that makes an argument ≤ 0, writing the rejection explicitly rather than silently dropping it.
Uses nPr to select a committee of 3 from 10, or nCr to arrange 5 distinct books in a row - choosing the formula by keyword rather than by asking whether order forms part of the outcome.
The two formulas look interchangeable and the wording is deceptive: 'select a president and secretary' sounds like selection but is ordered, while 'choose 3 toppings' is not. Identical objects and circular arrangements add further unmemorised adjustments.
What to do instead
Before computing, ask whether swapping two chosen items would produce a different outcome - if yes use permutations, if no use combinations. Memorise the refinements explicitly: divide by factorials for identical objects, use (n − 1)! for circular arrangements, and handle 'at least one' by complement.
Finds the term containing x⁵ in (2x − 3)⁸ by writing C(8, 5)(2x)⁵(−3)³, or drops the negative sign, or is off by one between r and r + 1 when asked for a specific term.
The index shift between 'term number' and r in the general term is a convention that is easy to misapply, and the sign of a negative second term depends on the parity of an exponent that is tempting to ignore when the arithmetic gets long.
What to do instead
Always write the general term before substituting: t_{r+1} = C(n, r) a^(n−r) b^r. Set the exponent of x equal to the target, solve for r, and only then substitute, keeping the minus sign inside b so the exponent handles the parity.