Calculus BC math practice

Every skill in this set

Limit at a removable discontinuityLimits & Continuity

factor, cancel the common factor, then substitute

The most-tested single limit computation, and the error is rarely the factoring — it is substituting before canceling, getting 0/0, and concluding the limit does not exist when it exists perfectly well.

Value making a piecewise function continuousLimits & Continuity

set the two branch expressions equal at the breakpoint

Continuity at a seam means the two branches must agree there. Students often differentiate first and match slopes, which is the wrong condition and gives the wrong constant.

Derivative evaluated at a pointDifferentiation: Definition & Rules

differentiate term by term, then substitute the x-value

A symbolic derivative cannot be graded numerically, but f′ at a point can — and that is what the exam asks for. The frequent error is substituting into f rather than into f′.

Derivative estimate from a tableDifferentiation: Definition & Rules

f′(a) ≈ [f(a + h) − f(a − h)] / (2h)

Table questions are guaranteed on the exam. The divisor is 2h, not h, and dividing by h is the single most common error — it doubles the answer and looks entirely reasonable.

Chain rule evaluated at a pointComposite & Implicit Differentiation

(f∘g)′(a) = f′(g(a)) · g′(a)

The error is almost never the rule — it is where each piece gets evaluated. f′ is evaluated at g(a), not at a, and losing that costs the whole question.

Derivative of an inverse functionComposite & Implicit Differentiation

g′(b) = 1 / f′(a), where f(a) = b and g = f⁻¹

Reciprocal is the easy half. The hard half is remembering to evaluate f′ at a, the original input, rather than at b — which produces a wrong number from a correctly recalled formula.

Linear approximation from a tangent lineContextual Applications

f(x) ≈ f(a) + f′(a)(x − a)

The tangent-line estimate appears every year. The error is dropping the (x − a) factor and adding the raw derivative, which treats a rate as a change.

Related rates: expanding circleContextual Applications

A = πr² ⟹ dA/dt = 2πr · dr/dt

Differentiating the area formula with respect to time is where the chain rule earns its keep. Forgetting the 2πr factor and reporting dr/dt as dA/dt is the classic slip.

Acceleration from a velocity polynomialContextual Applications

a(t) = v′(t); speeding up when v·a > 0

Speed and velocity are different quantities, and a negative acceleration only means slowing when velocity is positive. This drill forces the product test rather than a guess from signs.

Mean Value Theorem: finding cAnalytical Applications

f′(c) = [f(b) − f(a)] / (b − a)

Two computations that both go wrong: the average rate of change needs the division by (b − a), and then f′ must be set equal to it and solved — not evaluated at an endpoint.

Absolute extremum by the Candidates TestAnalytical Applications

compare f at every critical point and at both endpoints

The method is comparison of function values, and the endpoints are candidates. Students who evaluate only the critical points miss the answer whenever the extremum sits at an endpoint.

Definite integral of a polynomialIntegration & Accumulation

∫ₐᵇ f = F(b) − F(a)

An antiderivative cannot be graded numerically but a definite integral can, and the arithmetic is where marks are lost: substituting the lower limit and forgetting to subtract, or reversing the order.

Trapezoidal estimate with unequal widthsIntegration & Accumulation

each subinterval contributes [(f(x₀) + f(x₁))/2] · (x₁ − x₀)

Table data almost never has equal spacing, and the h/2 shortcut silently gives a wrong answer when it does not. Averaging heights and multiplying by that particular width always works.

Average value of a functionIntegration & Accumulation

average value = (1/(b − a)) ∫ₐᵇ f(x) dx

The division by (b − a) is the whole difficulty, and omitting it reports an accumulated total as an average. It is also easy to confuse with average rate of change, which involves no integral at all.

Amount from an initial value and a rateApplications of Integration

f(b) = f(a) + ∫ₐᵇ f′(t) dt

The integral of a rate is a CHANGE, not an amount. Reporting the change as the final quantity is the most common error on applied integration questions, because the initial value arrives in a different sentence.

Total distance versus displacementApplications of Integration

total distance = ∫|v(t)|dt, split at every sign change of v

Displacement and distance differ whenever the particle reverses, and integrating v without splitting lets the two directions cancel. The exam asks for distance specifically to test this.

Sum of a geometric seriesInfinite Sequences & Series

a/(1 − r), where a is the FIRST TERM PRESENT and r is the common ratio

The formula is memorized in the form that assumes the sum starts at n = 0. When it starts anywhere else, a is the first term that actually appears, and using the n = 0 term instead is the error this drill exists to catch.

Partial sum of a telescoping seriesInfinite Sequences & Series

write out the partial sum, cancel the interior terms, then take what is left

Series convergence is defined by the partial sums, and telescoping is the one family where you can find Sₙ exactly. Students who only ever apply convergence tests never practice the definition the tests are shortcuts for.

Derivative value from a Maclaurin coefficientInfinite Sequences & Series

the coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, so f⁽ⁿ⁾(0) = n! × coefficient

The exam asks for a high-order derivative that would be brutal to compute directly, and trivial to read off a series. Students who know the series but not the n! relationship report the coefficient itself as the derivative.

Radius of convergence from the ratio testInfinite Sequences & Series

set lim |aₙ₊₁/aₙ| < 1 and solve for |x|

A coefficient on x inside the power changes the radius, and students who have only seen Σxⁿ/bⁿ read the radius straight off the denominator without dividing by that coefficient.

Endpoint of an interval of convergenceInfinite Sequences & Series

center ± radius, then test each endpoint separately

The radius is the easy half. The interval needs the center applied in the right direction and each endpoint tested on its own, and a series can converge at one end and diverge at the other.

Taylor polynomial approximationInfinite Sequences & Series

P₃(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + f‴(a)(x − a)³/3!

Every term after the first carries a factorial and a power, and the two are the whole difficulty: dividing the third term by 3 instead of 2, or forgetting to raise (x − a) to the matching power, is the standard loss.

Alternating series error boundInfinite Sequences & Series

|S − Sₙ| ≤ the magnitude of the FIRST OMITTED term, aₙ₊₁

The bound is the next term, not the last one used. Off-by-one here is near-universal, and it produces a bound that is too large — which looks like a safe error and is still wrong.

Lagrange error boundInfinite Sequences & Series

|Rₙ(x)| ≤ M·|x − a|ⁿ⁺¹ / (n + 1)!

Both the exponent and the factorial are n + 1, not n — the bound reaches one order past the polynomial you actually used. Using n in either place is the mistake, and it changes the answer by a large factor.

Approximating with a known Maclaurin seriesInfinite Sequences & Series

cos x ≈ 1 − x²/2! + x⁴/4!

The four memorized series are only useful if the factorials and the parity of the exponents are right. Cosine has even powers with 2! and 4! underneath, and swapping in sine's odd powers is the failure this checks.

Slope of a parametric curveParametric, Polar & Vector-Valued Functions

dy/dx = (dy/dt) / (dx/dt)

The slope is a ratio of the two rates, and the order matters: dividing dx/dt by dy/dt gives the reciprocal, which is a plausible-looking number and the wrong one.

Second derivative of a parametric curveParametric, Polar & Vector-Valued Functions

d²y/dx² = [ d/dt (dy/dx) ] / (dx/dt)

The most-punished error in Unit 9 is computing (d²y/dt²)/(d²x/dt²), which is a different function entirely. The correct rule differentiates the slope expression with respect to t and then divides by dx/dt once more.

Speed of a particle in the planeParametric, Polar & Vector-Valued Functions

speed = |v(t)| = √[ (dx/dt)² + (dy/dt)² ]

Speed is the magnitude of the velocity vector, not either component and not their sum. The constant term in a position function also has to be differentiated away rather than carried into the answer.

Arc length of a parametric curveParametric, Polar & Vector-Valued Functions

L = ∫ₐᵇ √[ (dx/dt)² + (dy/dt)² ] dt

The integrand is the speed, so this is the same quantity as total distance traveled — and the setup fails in the same two ways: squaring after adding, or forgetting that the square root applies to the sum.

Position from a velocity vectorParametric, Polar & Vector-Valued Functions

x(t₁) = x(t₀) + ∫ from t₀ to t₁ of vₓ(t) dt

An integral of velocity gives the change in position, never the position. The initial coordinate has to be added back, and leaving it out is the same omission that breaks every net-change problem.

Area of one petal of a rose curveParametric, Polar & Vector-Valued Functions

A = (1/2)∫ from α to β of r² dθ, with α and β consecutive zeros of r

Two failures at once: the leading 1/2 gets dropped, and the limits are taken as 0 to 2π, which sweeps every petal instead of one. Both produce a clean number that is a whole-number multiple of the right answer.

Area enclosed by a limaçonParametric, Polar & Vector-Valued Functions

A = (1/2)∫₀^(2π) r² dθ, expanding r² before integrating

r² has to be expanded before integrating — squaring term by term is wrong, and the cross term is the one that vanishes over a full period while the cos² term does not.

Area between two polar curvesParametric, Polar & Vector-Valued Functions

A = (1/2)∫ from α to β of (r_outer² − r_inner²) dθ

The radii are squared first and then subtracted. Writing (r_outer − r_inner)² instead is the standard error, and it computes a quantity that is not an area at all.

Common questions