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Calculus 1 practice final (free, with answers)

A typical Calculus 1 final covers limits and continuity, the rules of differentiation, applications such as related rates, curve sketching and optimization, and a first look at integrals through the Fundamental Theorem of Calculus. Work each problem on paper before opening the answer, then compare your steps with the explanation.

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15 questions

Try each one before you open its answer.

  1. Limits and ContinuityMultiple choice

    1. Let f(x) = x² + k for x < 2 and f(x) = 3x − 1 for x ≥ 2. What value of k makes f continuous at x = 2?

    1. A k = −1
    2. B k = 1
    3. C k = 5
    4. D k = 9
    Show answer

    Answer: B. k = 1

    For continuity, the limit from the left must equal f(2). From the left, x² + k approaches 4 + k, and f(2) = 3(2) − 1 = 5, so 4 + k = 5 and k = 1.

  2. Limits and ContinuityShort answer

    2. Evaluate lim x→0 (1 − cos x) / x².

    Show answer

    Answer: 1/2

    Direct substitution gives the indeterminate form 0/0. One application of L'Hôpital's rule gives lim x→0 sin x / (2x), and since sin x / x → 1, the limit is 1/2.

  3. Derivative RulesMultiple choice

    3. What is the derivative of f(x) = x² / (x + 1)?

    1. A 2x
    2. B (3x² + 2x) / (x + 1)²
    3. C (x² + 2x) / (x + 1)²
    4. D (x² + 2x) / (x + 1)
    Show answer

    Answer: C. (x² + 2x) / (x + 1)²

    By the quotient rule, f′(x) = [(x + 1)(2x) − x² · 1] / (x + 1)² = (x² + 2x) / (x + 1)². Adding the two terms in the numerator instead of subtracting gives the wrong (3x² + 2x) / (x + 1)².

  4. Derivative RulesShort answer

    4. Find an equation of the tangent line to y = x³ − 4x + 1 at x = 2.

    Show answer

    Answer: y = 8x − 15

    At x = 2, y = 8 − 8 + 1 = 1. The slope is y′ = 3x² − 4, which equals 3(4) − 4 = 8 at x = 2, so the line is y − 1 = 8(x − 2), or y = 8x − 15.

  5. Chain Rule and Implicit DifferentiationMultiple choice

    5. If y = √(1 + 3x²), what is dy/dx?

    1. A 3x / √(1 + 3x²)
    2. B 1 / (2√(1 + 3x²))
    3. C 6x / √(1 + 3x²)
    4. D 3x · √(1 + 3x²)
    Show answer

    Answer: A. 3x / √(1 + 3x²)

    Write y = (1 + 3x²)^(1/2). The chain rule gives dy/dx = (1/2)(1 + 3x²)^(−1/2) · 6x = 3x / √(1 + 3x²). Forgetting to multiply by the inner derivative 6x gives 1 / (2√(1 + 3x²)) instead.

  6. Chain Rule and Implicit DifferentiationMultiple choice

    6. The curve x² + xy + y² = 7 passes through the point (1, 2). What is the slope dy/dx of the curve at that point?

    1. A −5/4
    2. B 4/5
    3. C −1
    4. D −4/5
    Show answer

    Answer: D. −4/5

    Differentiating both sides with respect to x, using the product rule on xy, gives 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + y) / (x + 2y). At (1, 2) this is −(2 + 2) / (1 + 4) = −4/5.

  7. Related RatesShort answer

    7. Air is pumped into a spherical balloon so that its volume increases at 100 cm³ per second. How fast is the radius increasing at the moment the radius is 5 cm? (Use V = (4/3)πr³.)

    Show answer

    Answer: 1/π cm per second, about 0.32 cm/s

    Differentiating with respect to time gives dV/dt = 4πr² · dr/dt. With dV/dt = 100 and r = 5, 100 = 4π(25) · dr/dt, so dr/dt = 100 / (100π) = 1/π ≈ 0.318 cm/s.

  8. Related RatesMultiple choice

    8. A 13-foot ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at 3 ft/s. How fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?

    1. A 3 ft/s
    2. B 7.2 ft/s
    3. C 1.25 ft/s
    4. D 0.8 ft/s
    Show answer

    Answer: C. 1.25 ft/s

    With x the distance from the wall to the bottom and y the height of the top, x² + y² = 169, so 2x(dx/dt) + 2y(dy/dt) = 0. When x = 5, y = 12, so dy/dt = −(5 × 3) / 12 = −1.25. The negative sign means the height is shrinking, so the top slides down at 1.25 ft/s.

  9. Curve SketchingMultiple choice

    9. Where does f(x) = x³ − 3x² − 9x + 5 have a local maximum and a local minimum?

    1. A Local maximum at x = −1, local minimum at x = 3
    2. B Local maximum at x = 3, local minimum at x = −1
    3. C Local maximum at x = 1, and no local minimum
    4. D Local minima at both x = −1 and x = 3
    Show answer

    Answer: A. Local maximum at x = −1, local minimum at x = 3

    f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), which is zero at x = −1 and x = 3. The derivative changes from positive to negative at x = −1, a local maximum, and from negative to positive at x = 3, a local minimum.

  10. Curve SketchingMultiple choice

    10. Where is f(x) = x⁴ − 4x³ concave down?

    1. A On (−∞, 0) and (2, ∞)
    2. B On (0, 2)
    3. C On (−∞, 3)
    4. D On (3, ∞)
    Show answer

    Answer: B. On (0, 2)

    f′(x) = 4x³ − 12x², so f″(x) = 12x² − 24x = 12x(x − 2). This is negative only between its zeros, so f is concave down on (0, 2) and concave up on (−∞, 0) and (2, ∞).

  11. OptimizationMultiple choice

    11. A farmer has 600 meters of fencing to enclose a rectangular field along a straight river, and no fence is needed along the river. What is the largest area the field can have?

    1. A 22,500 m²
    2. B 40,000 m²
    3. C 90,000 m²
    4. D 45,000 m²
    Show answer

    Answer: D. 45,000 m²

    Let x be each side perpendicular to the river, so the side along the river is 600 − 2x and the area is A = x(600 − 2x) = 600x − 2x². Then A′ = 600 − 4x = 0 at x = 150, which gives a 150 m by 300 m field with area 45,000 m².

  12. OptimizationShort answer

    12. An open-top box is made from a 12-inch by 12-inch square of cardboard by cutting equal squares from the four corners and folding up the sides. What is the largest possible volume of the box?

    Show answer

    Answer: 128 cubic inches, from cutting 2-inch squares

    With corner squares of side x, V = x(12 − 2x)² for 0 < x < 6. V′ = (12 − 2x)(12 − 6x), which is zero inside that interval only at x = 2, and V(2) = 2 · 8² = 128 cubic inches, while V approaches 0 at both ends of the interval.

  13. Definite Integrals and the FTCMultiple choice

    13. Let F(x) = ∫₁^(x²) √(1 + t³) dt. What is F′(x)?

    1. A √(1 + x⁶)
    2. B 2x√(1 + x⁶)
    3. C √(1 + x³)
    4. D 2x√(1 + x⁶) − √2
    Show answer

    Answer: B. 2x√(1 + x⁶)

    By the Fundamental Theorem of Calculus with the chain rule, F′(x) = √(1 + (x²)³) · d/dx(x²) = 2x√(1 + x⁶). The constant lower limit adds nothing to the derivative.

  14. Definite Integrals and the FTCShort answer

    14. Find the area of the region enclosed by the curves y = x² and y = 2x.

    Show answer

    Answer: 4/3

    The curves meet where x² = 2x, at x = 0 and x = 2, and y = 2x is on top between them. The area is ∫₀² (2x − x²) dx = [x² − x³/3] from 0 to 2 = 4 − 8/3 = 4/3.

  15. SubstitutionMultiple choice

    15. Evaluate the indefinite integral ∫ x e^(x²) dx.

    1. A e^(x²) + C
    2. B 2x e^(x²) + C
    3. C (1/2) e^(x²) + C
    4. D (x²/2) e^(x²) + C
    Show answer

    Answer: C. (1/2) e^(x²) + C

    Let u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫ eᵘ du = (1/2)eᵘ + C = (1/2) e^(x²) + C, which you can check by differentiating.

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