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

Curves with no breaks

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

  1. If the limit of f(x) as x approaches 5 exists and equals f(5), then f is continuous at x = 5.

    Circle one:   True   False

  2. A function f is defined at x = 3. Which of these is NOT one of the three things required for f to be continuous at x = 3?

    • f(3) has a value, so the function is defined there
    • the limit of f(x) as x approaches 3 exists
    • the derivative f'(3) exists
    • the limit of f(x) as x approaches 3 equals f(3)

  3. Which of these is NOT needed for continuity at x = 3?

    • f(3) is defined
    • The limit at 3 exists
    • The derivative at 3 exists

  4. What kind of break does f(x) = 1 / x have at x = 0?

    • Infinite
    • Removable
    • Jump

  5. A function follows f(x) = x + 1 when x is less than 2, and f(x) = 5 when x is 2 or more. What kind of discontinuity is at x = 2?

    • removable
    • jump
    • infinite
    • none, it is continuous

  6. f(x) = x squared - 2 is continuous on [1, 2] with f(1) = -1 and f(2) = 2. At least how many roots sit inside (1, 2)?

    Answer: ______________

  7. The function f(x) = (x squared minus 9) / (x minus 3) is not defined at x = 3. What kind of discontinuity does it have at x = 3?

    • removable
    • jump
    • infinite
    • none, it is continuous

  8. f(x) = x squared minus 2 is continuous on [1, 2], with f(1) = -1 and f(2) = 2. According to the Intermediate Value Theorem, at least how many roots does f have in the interval (1, 2)?

    Answer: ______________

  9. Which situation means the Intermediate Value Theorem does NOT guarantee a root on [a, b]?

    • f is continuous on [a, b] and f(a) is negative while f(b) is positive
    • f(a) and f(b) have opposite signs but f has a jump discontinuity inside the interval
    • f is continuous on [a, b] and f(a) = f(b) = 0
    • f is continuous on [a, b] and f(a) is positive while f(b) is positive but the graph dips below zero in the middle

  10. For f(x) = x cubed + x - 1, f(0) = -1 and f(1) = 1. What licenses the claim of a root between 0 and 1?

    • Continuity plus the sign change, through the Intermediate Value Theorem
    • Factoring pinpoints the root exactly
    • The cubic must cross three times
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Answer key

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

Curves with no breaks W1-mt_kaQR6cWjbV-s1

  1. True · That one sentence quietly includes all three conditions: f(5) must be defined for the statement to make sense, the limit exists, and the limit equals the value.
  2. the derivative f'(3) exists · Continuity needs only three things: the function is defined at the point, the limit exists there, and the two agree. Needing a derivative is a stronger, different condition.
  3. The derivative at 3 exists · Continuity needs defined value, existing limit and agreement. Slopes are extra.
  4. Infinite · Values near 0 blow up in size instead of settling.
  5. jump · The left side approaches 3 while the right side equals 5. Both one-sided limits exist but they disagree, so the graph leaps upward at x = 2.
  6. 1 · A sign change on an unbroken curve forces at least one crossing.
  7. removable · The numerator factors as (x minus 3)(x plus 3), so the troublesome factor cancels. The limit exists but the point itself is missing, which is a removable discontinuity.
  8. 1 · A sign change from -1 to 2 on a continuous function forces the graph to cross the x-axis at least once between x = 1 and x = 2.
  9. f(a) and f(b) have opposite signs but f has a jump discontinuity inside the interval · The theorem requires continuity on the whole closed interval. A jump discontinuity breaks the unbroken curve the theorem relies on, so no root is guaranteed even with a sign change.
  10. Continuity plus the sign change, through the Intermediate Value Theorem · A polynomial is unbroken, and opposite ends force a crossing.
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