The Mean Value Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

One instant matches the average

Mathematics · Calculus & Analysis · ages 18-19
Name ______________________   Date ____________

  1. For f(x) = 2x^2, compute the average rate of change over the interval [1, 3].

    Answer: ______________

  2. For f of x equals 2 x squared, find the average rate of change over 1 to 3.

    Answer: ______________

  3. For g of x equals x squared minus 4 x plus 3 on 1 to 3, all conditions hold and the endpoint values match. Which c is guaranteed?

    • c equals 2
    • c equals 1
    • c equals 3

  4. The theorem needs differentiability on the closed interval and continuity only on the open interval.

    Circle one:   True   False

  5. You try to apply Rolle's theorem to f(x) = |x - 2| on the interval [0, 4]. The endpoint values are f(0) = 2 and f(4) = 2, so they match. Which hypothesis still fails?

    • The function is not continuous on [0, 4]
    • The function is not differentiable at x = 2
    • The endpoint values are not equal
    • No hypothesis fails, so the theorem must hold

  6. Rolle's theorem is the special case of the Mean Value Theorem where the endpoint values match. For g(x) = x^2 - 4x + 3 on [1, 3], the hypotheses all hold. What value of c in (1, 3) does the theorem guarantee?

    • c = 1
    • c = 2
    • c = 3
    • c = 4

  7. For the function f(x) = x - 3/x on the interval [1, 3], the Mean Value Theorem promises a number c inside (1, 3) where the instantaneous rate equals the average rate. Find that c. Round your answer to two decimal places.

    Answer: ______________

  8. You want a function that is continuous everywhere but fails the differentiability hypothesis of the Mean Value Theorem at x = 0. Which one works?

    • f(x) = x^2
    • f(x) = |x|
    • f(x) = x^3
    • f(x) = 2x + 1

  9. Suppose f of 1 is 2 and the derivative is at most 5 everywhere. What is the largest possible value of f of 4?

    Answer: ______________

  10. Suppose f(1) = 2 and the derivative satisfies f'(x) is at most 5 everywhere. Using the Mean Value Theorem, what is the largest possible value of f(4)?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

One instant matches the average W1-mt_09iApYVelx-s1

  1. 8 · The average rate of change is (f(3) - f(1)) / (3 - 1) = (18 - 2) / 2 = 8. This secant slope is what the Mean Value Theorem matches with a tangent somewhere inside.
  2. 8 · 18 minus 2 over 3 minus 1 is 8.
  3. c equals 2 · Both endpoint values are 0, and g prime of x is 2 x minus 4, so c equals 2.
  4. False · It is the other way round: continuity on the closed interval, differentiability on the open one.
  5. The function is not differentiable at x = 2 · The absolute value graph has a sharp corner at x = 2, where the slope jumps from -1 to 1, so no derivative exists there even though the function is continuous.
  6. c = 2 · g(1) = 0 and g(3) = 0, and a polynomial is continuous and differentiable everywhere, so Rolle's theorem applies and gives c = 2.
  7. 1.73 · The average rate over [1, 3] is 2, and solving f'(c) = 1 + 3/c^2 = 2 gives c = sqrt(3), which is about 1.73 and lies inside the interval.
  8. f(x) = |x| · |x| has a sharp corner at 0 where the slope jumps from -1 to 1, so no derivative exists there, yet the graph has no gaps and stays continuous.
  9. 17 · f of 4 equals 2 plus 3 times f prime of c, at most 2 plus 15.
  10. 17 · The Mean Value Theorem gives a c in (1, 4) with f(4) - f(1) = f'(c) * 3, and since f'(c) is at most 5, f(4) is at most 2 + 15 = 17.
Worksheet · LightMySky