Before calculus can even begin, AP Calculus AB leans heavily on a solid command of functions themselves — how to find their zeros, invert them, determine their domain and range, and recognize periodic and logarithmic behavior. This foundational material shows up constantly throughout the course: understanding inverse functions is essential for differentiating inverse trig functions later, recognizing the range of a composite function matters when analyzing exponential growth models, and comfort with polynomial zeros underlies almost every optimization and related-rates problem on the exam.
A recurring skill tested here is function composition — substituting one function's output into another, whether that's evaluating f(2x) once you already know the zeros of f(x), confirming that f(g(x)) = x for a function and its inverse, or determining how tightly a composed function's range is constrained by a restricted domain. Inverse functions deserve special attention: finding f−1(x) means solving the equation y = f(x) for x in terms of y, and it's worth double-checking the result by confirming that f and f−1 really do undo each other. Logarithmic and exponential equations often hide an ordinary algebra problem behind unfamiliar notation, and periodicity questions reward knowing the standard period formulas for sine and cosine cold.
Below are 12 AP Calculus AB-style practice problems covering zeros of polynomials, inverse functions, reflections and transformations, periodicity, domain and range, and logarithmic equations. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.
If the solutions to the equation g(x) = 0 are x = 3, −4, then g(2x) = 0 at x =
g(2x) = 0 means the input to g, which is 2x, must equal one of g's known zeros, 3 or −4.
Set 2x = 3: x = 3/2. Set 2x = −4: x = −2.
The correct answer is B. It's tempting to double the original zeros instead of halving them — but compressing the graph horizontally (as 2x does) means the zeros occur at smaller x-values, not larger ones.
The set of zeros of f(x) = x3+6x2+9x is
Factor out the common factor x first: f(x) = x(x2+6x+9).
The quadratic factor is a perfect square trinomial: x2+6x+9 = (x+3)2. So f(x) = x(x+3)2.
Setting each factor equal to zero gives x = 0 and x = −3 (the −3 root is repeated, but the set of distinct zeros only lists it once).
The correct answer is B.
The values of x for which the graphs of y = x+3 and y2 = 4x intersect are
Substitute y = x+3 into y2 = 4x: (x+3)2 = 4x.
Expand and simplify: x2+6x+9 = 4x → x2+2x+9 = 0.
Check the discriminant: b2−4ac = 4 − 36 = −32, which is negative — so there is no real solution for x.
The correct answer is D, no intersection.
Keep the momentum going.
Explore 1,500+ Free AP Calculus AB ProblemsThe function whose graph is a reflection in the y-axis of the graph of f(x) = 2−5x is
A reflection in the y-axis replaces every x in the original function with −x: g(x) = f(−x).
Substitute −x for x in f(x) = 2−5x: g(x) = 2−5−x.
The correct answer is A. Choices C and D use logarithms, which would reflect the graph across the line y=x (an inverse), not across the y-axis — a common mix-up between the two transformations.
Let h(x) have an inverse function k(x). Then h(k(x)) =
By definition, an inverse function undoes the original function. Applying h and then its inverse k (or k then h) always returns the original input.
h(k(x)) = x for every x in the domain of k, and likewise k(h(x)) = x for every x in the domain of h.
The correct answer is B.
The function f(x) = x3+2x−30 has exactly one real zero. It is between
Since f is a polynomial (continuous everywhere), a sign change between two x-values guarantees a zero somewhere in between. Test the boundary values of each choice:
f(2) = 8+4−30 = −18 and f(3) = 27+6−30 = 3.
f changes from negative to positive between x=2 and x=3, so a zero must exist somewhere in that interval. (Checking the other intervals confirms f stays negative throughout each of them, with no sign change.)
The correct answer is D.
6 down, 6 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB ProblemsThe period of f(x) = sin((3π/2)x) is
For f(x) = sin(bx), the period is given by 2π/b. Here, b = 3π/2.
period = 2π / (3π/2) = 2π · (2/(3π)) = 4/3.
The correct answer is A.
The range of y = f(x) = ln(sin x), for 0 < x < π, is
On the interval 0 < x < π, sin x takes every value in (0, 1] — it never reaches 0 (the endpoints are excluded) but it does reach its maximum of 1 (at x=π/2).
Apply ln to that entire range. As the input approaches 0 from the right, ln(sin x) → −∞. At sin x = 1, ln(1) = 0, which is attained.
So the range of ln(sin x) is every real number up to and including 0: (−∞, 0].
The correct answer is A.
If logb(5b) = b/2, then b =
Use the power rule for logarithms to pull the exponent out front: logb(5b) = b · logb(5).
Set this equal to the given value: b · logb(5) = b/2. Since b ≠ 0, divide both sides by b: logb(5) = 1/2.
Rewrite in exponential form: b1/2 = 5, which means √b = 5.
Square both sides: b = 25. The correct answer is D.
9 down, 3 to go.
Explore 1,500+ Free AP Calculus AB ProblemsLet f−1 be the inverse function of f(x) = x3−5. Then f−1(x) =
To find an inverse, write y = f(x), then swap x and y and solve for the new y. Start with y = x3−5.
Swap variables: x = y3−5. Solve for y: y3 = x+5.
Take the cube root of both sides: y = ∛(x+5).
The correct answer is C.
The set of x-intercepts of the graph of f(x) = x3−2x2−5x+6 is
Test small integer values to find a root. f(1) = 1−2−5+6 = 0 — so x=1 is a root, and (x−1) is a factor.
Dividing f(x) by (x−1) gives the quotient x2−x−6, which factors further as (x−3)(x+2).
So f(x) = (x−1)(x−3)(x+2), with zeros at x = 1, 3, −2.
The correct answer is C, {−2, 1, 3}.
If the domain of f is restricted to the open interval (0, π), then the range of f(x) = ecot x is
On the interval (0, π), cot x is continuous and strictly decreasing, moving from +∞ (as x → 0+) down to −∞ (as x → π−). So cot x takes on every real number exactly once on this interval.
Since the exponent cot x ranges over all real numbers, and eu takes every positive value as u ranges over all reals (but never reaches 0 or goes negative), the output ecot x takes every positive real value.
The correct answer is B, the set of positive reals.
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