Continuity and the Intermediate Value Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Mathematics · Calculus & Analysis · ages 18-19
Name ______________________   Date ____________
  1. Continuity at the left endpoint of a closed interval uses the right-hand limit available within that interval.

    Circle one:   True   False

  2. At an interior point 2, f(2) = 9 and the finite limit is 7. Which continuity requirement fails?

    • The function value must be defined.
    • The finite limit must exist.
    • The limit must equal the value.
    • The one-sided limits must be finite.
  3. Which condition is NOT required for continuity at the interior point 3?

    • f(3) is defined
    • The finite limit at 3 exists
    • The derivative at 3 exists
  4. What kind of discontinuity does 1/x have at 0?

    • Infinite
    • Removable
    • Jump
  5. Let f(x) = 2x below 3 and f(x) = x+k at or above 3. Which k makes f continuous at 3?

    • 6
    • 3
    • 0
    • 9
  6. For f(x) = x²-2 on [1,2], f(1) = -1 and f(2) = 2. At least how many roots does the Intermediate Value Theorem guarantee in (1,2)?

    Answer: ______________

  7. At -1, a function has finite limit 4 but assigned value 10. Which replacement value makes it continuous there?

    • 4
    • 10
    • 6
    • 14
  8. A continuous function on [0,2] has f(0) = -2, f(1) = 3 and f(2) = -1. At least how many distinct interior roots are guaranteed?

    Answer: ______________

  9. For f(x) = x(1-x) on [0,1], both endpoint values are zero and f is positive inside. Which conclusion is correct?

    • Two endpoint roots force a third root inside.
    • The endpoint roots do not establish an interior root.
    • The polynomial is discontinuous because its endpoint values agree.
    • The Intermediate Value Theorem forces zero at the midpoint.
  10. For f(x) = x³+x-1, f(0) = -1 and f(1) = 1. What justifies an interior root?

    • Polynomial continuity and opposite endpoint signs.
    • The degree alone fixes an exact root between the endpoints.
    • Opposite signs locate the root at the interval's midpoint.
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Answer key

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

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  1. True · Inputs within the interval approach its left endpoint from the right.
  2. The limit must equal the value. · Both the value and finite limit exist, but 7 differs from 9.
  3. The derivative at 3 exists · Continuity requires the value, finite limit and their equality, not a derivative.
  4. Infinite · Outputs grow without bound in magnitude near the vertical asymptote.
  5. 3 · Continuity requires 6 = 3+k, giving k = 3; the assigned right-branch value then also equals 6.
  6. 1 · Continuity and opposite endpoint signs guarantee at least one interior root.
  7. 4 · The point value must equal the existing finite limit 4.
  8. 2 · There is a sign change on each of [0,1] and [1,2], giving roots in disjoint open intervals.
  9. The endpoint roots do not establish an interior root. · The endpoints are roots, but positivity on (0,1) rules out an interior root.
  10. Polynomial continuity and opposite endpoint signs. · The polynomial is continuous and zero is strictly between its endpoint outputs.
Worksheet · LightMySky