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Introduction to Integrals · Ages 17–18

Finding a Function from Its Derivative and a Point

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For 12th grade calculus students. Find the original function f(x) or position function s(t) using the derivative and given point.

  1. Find f(x) if f'(x) = 4 and the curve passes through (1, 6).

  2. Find f(x) if f'(x) = 2x - 5 and f(0) = 4.

  3. Find f(x) if f'(x) = 3x^2 + 2x and the graph passes through (2, 10).

  4. Find f(x) if f'(x) = 6x^2 - 4x + 3 and f(1) = 8.

  5. A particle moves with velocity v(t) = 12t^3 - 2t and position s(1) = 7. Find s(t).

  6. Find f(x) if f'(x) = 1 / x^2 (for x > 0) and the curve passes through (2, 1).

  7. Find f(x) given f'(x) = 3x^2 - 2, and the tangent line at x = 2 is y = 10x - 17.

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Finding a Function from Its Derivative and a Point

Answer key

  1. Find f(x) if f'(x) = 4 and the curve passes through (1, 6).

    f(x) = 4x + 2

  2. Find f(x) if f'(x) = 2x - 5 and f(0) = 4.

    f(x) = x^2 - 5x + 4

  3. Find f(x) if f'(x) = 3x^2 + 2x and the graph passes through (2, 10).

    f(x) = x^3 + x^2 - 2

  4. Find f(x) if f'(x) = 6x^2 - 4x + 3 and f(1) = 8.

    f(x) = 2x^3 - 2x^2 + 3x + 5

  5. A particle moves with velocity v(t) = 12t^3 - 2t and position s(1) = 7. Find s(t).

    s(t) = 3t^4 - t^2 + 5

  6. Find f(x) if f'(x) = 1 / x^2 (for x > 0) and the curve passes through (2, 1).

    f(x) = -1/x + 3/2

  7. Find f(x) given f'(x) = 3x^2 - 2, and the tangent line at x = 2 is y = 10x - 17.

    The point of tangency is (2, 3). Integrating gives f(x) = x^3 - 2x + C. Substituting (2, 3) yields C = -1, so f(x) = x^3 - 2x - 1.