Calculus AB 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.

Common questions