CLEP Precalculus common mistakes
CLEP Precalculus · Understand the traps and how to avoid them.
When solving an equation such as 1/(x-3) + 2 = x/(x-3), the candidate multiplies both sides by (x-3) and reports the resulting solution x = 3 without checking it, even though x = 3 makes the original denominators zero.
Multiplying by an expression containing the variable can introduce extraneous solutions, but candidates treat clearing denominators as a purely mechanical step and skip the final substitution check because the algebra 'worked out'.
What to do instead
Before solving, write down the restricted values (every value making any denominator zero). After solving, compare each candidate against that list and substitute back into the original equation. Practice by deliberately constructing equations whose cleared form has a root equal to a restricted value, so the check becomes automatic.
The candidate solves |2x - 5| < 7 by writing only 2x - 5 < 7 and getting x < 6, dropping the lower bound and reporting an unbounded ray instead of the interval (-1, 6).
Absolute value is treated as 'just remove the bars', and the compound nature of the inequality is not internalized; the negative case feels redundant because the positive case already produced an answer.
What to do instead
Rewrite every absolute-value inequality explicitly as a compound statement first: |A| < b becomes -b < A < b, and |A| > b becomes A < -b or A > b. Then solve each piece and intersect (for 'less than') or union (for 'greater than'). Always test one point inside and one outside the claimed solution set.
Asked for the domain of f(x) = sqrt(x - 2)/(x - 5), the candidate gives all reals, or gives x >= 2 while forgetting to exclude x = 5.
Candidates check only one restriction type at a time; radicals and denominators are learned in separate chapters, so when both appear in one expression only the most recently studied restriction gets applied.
What to do instead
Adopt a fixed checklist applied to every domain question: (1) even roots require radicand >= 0, (2) denominators require nonzero, (3) log arguments require > 0. Solve each condition separately, then intersect. State the answer in interval notation and verify with two test numbers, one that should be excluded and one that should be included.
Given f(x) = x^2 and g(x) = x - 3, the candidate computes (f o g)(x) as x^2 - 3 instead of (x - 3)^2, substituting into the wrong function or distributing the composition as if it were multiplication.
The notation f(g(x)) is read left-to-right as 'f then g', and the visual similarity to f(x)·g(x) invites treating composition as a product.
What to do instead
Read (f o g)(x) aloud as 'f of g of x' and physically write f( ) with a blank, then place the entire expression for g(x) into the blank with parentheses: f(x - 3) = (x - 3)^2. Verify numerically with a value like x = 5: compute g(5) = 2, then f(2) = 4, and confirm the symbolic form gives 4.
When asked whether a graphed or tabulated relation is a function, the candidate answers yes because 'it looks like a curve', or applies the vertical line test to a graph of the inverse situation, without checking whether any input value repeats with different outputs.
The definition of function as 'each input has exactly one output' is memorized as a slogan but not operationalized; candidates rely on visual impression rather than testing repeated x-values.
What to do instead
Convert the question into a concrete test: scan the table or graph for two points sharing the same x but different y. If any exist, it is not a function. For graphs, state the vertical line test explicitly and name a specific vertical line that fails. Practice with relations given as tables, ordered-pair sets, and graphs so the same test is applied in all three representations.
The candidate transforms y = f(x) into y = f(x - 4) + 2 and describes the shift as 'left 4, up 2', reversing the horizontal direction.
Horizontal transformations act opposite to intuition: subtracting inside the argument moves the graph right, but the minus sign suggests left. Candidates memorize 'minus means left' from number-line habits.
What to do instead
Anchor the rule to a tracked point: if the new formula is f(x - h) + k, the point (0, f(0)) moves to (h, f(0) + k). Compute h directly from the expression rather than from the sign's appearance. For each transformation problem, identify the image of one or two specific key points and confirm the direction before writing the description.
Asked for the center and radius of x^2 + y^2 - 6x + 4y - 12 = 0, the candidate reads the center as (-6, 4) or the radius as sqrt(12), skipping the completion of the square.
The standard form (x - h)^2 + (y - k)^2 = r^2 is recognized, but the general form's coefficients are mistaken for the center, and completing the square is a multi-step procedure candidates avoid under time pressure.
What to do instead
Always convert general form to standard form: group x terms and y terms, complete the square by adding (b/2)^2 to both sides, and factor. Remember the center coordinates are the opposites of the numbers subtracted inside the parentheses, and the radius is the square root of the right-hand side. Check by expanding the standard form back to the original equation.
The candidate evaluates sin(150°) as -sin(30°) = -1/2 because 150° is in the second quadrant, or evaluates cos(210°) as +sqrt(3)/2, getting the sign pattern wrong.
Reference angles are computed correctly but the quadrant sign rule is applied from memory without a consistent scheme; candidates confuse 'sine is negative in quadrant II' with 'angles past 90° are negative'.
What to do instead
Draw the angle in standard position, mark the reference angle to the x-axis, and use the mnemonic 'All Students Take Calculus' (quadrant I all positive, II sine, III tangent, IV cosine) to assign the sign before computing the magnitude. Write the sign and the magnitude as two separate steps so an error in one does not hide the other.
Solving sin(x) = 1/2 on [0, 2π), the candidate gives only x = π/6 and omits x = 5π/6; similarly, solving tan(x) = 1 the candidate gives only π/4 and misses π/4 + π.
Inverse-function thinking produces a single principal value, and the periodic symmetry of the unit circle is not systematically applied; candidates stop as soon as one solution is found.
What to do instead
After finding the reference solution, use the unit circle to locate every angle in the requested interval with the same sine, cosine, or tangent value: for sine, reflect across the y-axis; for cosine, across the x-axis; for tangent, add π. Then list solutions in order and count them against the interval. Practice by graphing the function and the horizontal line to see all intersections.
Given a word problem about a population growing 4% per year from 12,000, the candidate writes P(t) = 12000(4)^t or P(t) = 12000 + 0.04t instead of 12000(1.04)^t, and then answers a 'when will it double' question by linear extrapolation.
Percent rates are confused with the growth factor, and the distinction between linear and exponential change is not tied to the phrase 'percent of the current amount each year'.
What to do instead
For any model, first identify the initial value a and the multiplier b: for a percent increase r, b = 1 + r; for a decrease, b = 1 - r. Write the model as y = a·b^t and sanity-check with t = 1 and t = 2 by hand. For doubling questions, set a·b^t = 2a, cancel a, and solve with logarithms rather than estimating linearly.