AP Calculus BC Practice: Functions (10 Step-by-Step Explanations)

AP Calculus BC · Functions

Functions: AP Calculus BC Practice Problems

AP Calculus BC expects a comfortable, calculator-active command of functions in every form they can appear: explicit functions built from exponentials, trig, and radicals; functions defined parametrically as x(t) and y(t); and functions defined in polar coordinates as r(θ). Many BC questions on this material aren't really testing new calculus concepts — they're testing whether you can set up an equation correctly and then solve it numerically, since most of these functions don't have a clean algebraic root or intersection point.

For ordinary functions, common tasks include finding an x-intercept when the function only crosses once (a straightforward numerical root), finding an x-intercept on a specific piece of the graph when a function crosses multiple times and only one crossing matches an extra condition like "where f is decreasing," identifying every interval where a function is negative (which usually means finding all the roots first, then testing the sign in between them), counting how many times an oscillating function crosses the x-axis, and finding the domain of an inverse function (which is just the range of the original). For parametric and polar curves, the recurring skills are eliminating the parameter to identify what kind of curve is being described, converting between a parametric or polar equation and a rectangular one, finding where a polar curve passes through the origin, and finding where two curves given in polar form intersect.

Below are 10 AP Calculus BC-style practice problems covering all of these skills. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation, including the exact equation to solve numerically where a calculator is required. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus BC QBank.

Finding a single x-intercept numerically

The graph of the function f(x) = 3e^cos(x)−4 crosses the x-axis exactly once in the interval [1,2]. What is the x-coordinate of this x-intercept?

Full Explanation

This equation can't be solved algebraically, so use a graphing calculator to find where 3e^cos(x) = 4 on [1,2].

Checking the endpoints confirms a single crossing: f(1) ≈ 1.150 > 0 and f(2) ≈ −2.021 < 0, and since cos(x) is monotonic on this interval, there's exactly one root.

Solving numerically gives x ≈ 1.279.

The correct answer is B.

Finding the x-intercept on a specific piece of the graph

Find the x-intercept of the graph of f(x) = √(cos(x)+1) + x³ − 4e^sin(x) + 7 on the portion of the graph where f(x) is decreasing.

Full Explanation

A calculator search over a wide window shows f has three x-intercepts: approximately −1.849, 0.984, and 1.367.

Since the question asks specifically for the intercept where f is decreasing, check the sign of f′ at each candidate (either graphically, by seeing where the curve is sloping downward as it crosses, or by evaluating f′ numerically).

Only at x ≈ 0.984 is f′(x) < 0 — the other two crossings occur while f is increasing.

The correct answer is C.

Finding all intervals where a function is negative

You are given the function f(x) = e^{x⁴−3x²+1} − 6 on the closed interval [−2,2]. Find all intervals where f(x) < 0.

Full Explanation

Since x⁴−3x²+1 is an even function (only even powers of x), f itself is symmetric about the y-axis, so its zeros come in a ± pair.

Solving e^{x⁴−3x²+1} = 6 numerically gives x ≈ ±1.801.

Test a point between the roots, like x=0: f(0) = e¹−6 ≈ −3.28 < 0. Test a point outside, like x=2: f(2) = e⁵−6 ≈ 142.4 > 0.

So f(x) < 0 on the single interval between the two roots: (−1.801, 1.801). The correct answer is B.

Keep the momentum going.

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Counting x-axis crossings of an oscillating function

You are given the function f(x) = (6−3x−2x²)cos(2x−1) on the closed interval [−3,2]. How many times does f(x) cross the x-axis in the interval?

Full Explanation

f(x) is a product of two factors, so it crosses zero wherever either factor crosses zero: the quadratic 6−3x−2x², or the oscillating cos(2x−1).

The quadratic has two real roots (found with the quadratic formula), both inside [−3,2].

The cosine factor completes several oscillations over this interval (since its period is 2π/2 = π ≈ 3.14, and the interval has width 5), producing several more crossings.

Graphing (or tracking sign changes numerically) across the full interval counts five total crossings. The correct answer is B.

Finding a coordinate on a polar curve

On the interval [0,2π], there is one point on the curve r = θ−2sinθ whose x-coordinate is 1.5. Find the y-coordinate there.

Full Explanation

Convert to rectangular coordinates: x(θ) = r cosθ = (θ−2sinθ)cosθ and y(θ) = r sinθ = (θ−2sinθ)sinθ.

Use a calculator to solve x(θ) = 1.5 for θ on [0,2π]. This gives θ ≈ 4.932.

Substitute this θ into the y(θ) formula: y(4.932) ≈ −6.719.

The correct answer is A.

Domain of an inverse function

If f(x) = 2+e^x, then the domain of f⁻¹(x) is

Full Explanation

The domain of f⁻¹ is exactly the range of f, since an inverse function swaps the roles of input and output.

Since e^x > 0 for every real x, we always have f(x) = 2+e^x > 2, and f(x) can be made arbitrarily close to 2 (as x→−∞) or arbitrarily large (as x→∞).

So the range of f is (2, ∞), which becomes the domain of f⁻¹.

The correct answer is C.

6 down, 8 to go — keep practicing.

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Identifying a curve from its parametric equations

The curve defined parametrically by x(t) = t²+5 and y(t) = t²+2 is a part of a(n)

Full Explanation

Eliminate the parameter t by solving for it in one equation and substituting into the other. From x = t²+5: t² = x−5.

Substitute into the equation for y: y = t²+2 = (x−5)+2 = x−3.

This is a linear relationship between x and y (with slope 1), so the curve is part of a line — specifically the ray where x≥5, since t²≥0 restricts x.

The correct answer is A.

Converting a parametric curve to a rectangular equation

Which equation includes the curve defined parametrically by x(t) = cos²(t) and y(t) = 3sin(t)?

Full Explanation

From the given equations, cos²(t) = x and sin(t) = y/3, so sin²(t) = y²/9.

Use the Pythagorean identity cos²(t)+sin²(t) = 1: x + y²/9 = 1.

Multiply both sides by 9 to clear the fraction: 9x + y² = 9.

The correct answer is C.

Finding where a polar rose passes through the origin

Find the smallest value of θ in the interval [0,2π] for which the rose r = 3cos(7θ) passes through the origin.

Full Explanation

The curve passes through the origin whenever r=0: 3cos(7θ) = 0 → cos(7θ) = 0.

Cosine is first zero (for a positive argument) at 7θ = π/2, so θ = π/14.

Since θ=0 gives r=3 (not zero), the smallest θ producing r=0 is indeed π/14.

The correct answer is C.

9 down, 5 to go.

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Intersection of two polar curves

For what value of θ in the interval [0,π] do the polar curves r = 1 and r = 4+6cosθ intersect?

Full Explanation

At an intersection, both curves have the same r-value for the same θ, so set the two expressions equal: 1 = 4+6cosθ.

Solve for cosθ: 6cosθ = −3 → cosθ = −1/2.

On the interval [0,π], cosθ = −1/2 at θ = 2π/3.

The correct answer is D.

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