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Quiz: Mean Value Theorem and Extrema

Test your understanding of the Mean Value Theorem and extrema with these review questions.

1. The Mean Value Theorem requires a function to be:

  1. Continuous on [a,b] only
  2. Differentiable on (a,b) only
  3. Continuous on [a,b] and differentiable on (a,b)
  4. Differentiable everywhere
Show Answer

The correct answer is C. The MVT requires two conditions: continuity on the closed interval [a,b] AND differentiability on the open interval (a,b). Both are necessary.

Concept Tested: MVT Conditions

2. The Mean Value Theorem guarantees the existence of a point c where:

  1. f(c) = 0
  2. f'(c) = 0
  3. f'(c) = [f(b) − f(a)]/(b − a)
  4. f(c) = [f(a) + f(b)]/2
Show Answer

The correct answer is C. The MVT says there exists c in (a,b) where f'(c) equals the average rate of change over [a,b]. Geometrically, the tangent line is parallel to the secant line.

Concept Tested: MVT Conclusion

3. Rolle's Theorem is a special case of the MVT when:

  1. f(a) = 0
  2. f(b) = 0
  3. f(a) = f(b)
  4. f'(a) = f'(b)
Show Answer

The correct answer is C. Rolle's Theorem applies when f(a) = f(b). The conclusion is that f'(c) = 0 for some c in (a,b)—a horizontal tangent must exist somewhere between.

Concept Tested: Rolle's Theorem

4. The Extreme Value Theorem guarantees that a continuous function on [a,b]:

  1. Has a derivative everywhere
  2. Attains both a maximum and minimum value
  3. Has a zero somewhere in (a,b)
  4. Is differentiable on (a,b)
Show Answer

The correct answer is B. The Extreme Value Theorem states that a continuous function on a closed interval must attain both a global maximum and a global minimum value on that interval.

Concept Tested: Extreme Value Theorem

5. A critical number of f(x) is a value c in the domain where:

  1. f(c) = 0
  2. f'(c) = 0 or f'(c) does not exist
  3. f(c) is a maximum or minimum
  4. f is continuous
Show Answer

The correct answer is B. A critical number is where f'(c) = 0 or f'(c) does not exist (and c is in the domain). Critical numbers are candidates for local extrema.

Concept Tested: Critical Number

6. To find the absolute maximum of f(x) on [a,b], you should:

  1. Find where f'(x) = 0 only
  2. Evaluate f at critical numbers and endpoints, then compare
  3. Find where f''(x) < 0
  4. Graph the function
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The correct answer is B. The Candidates Test: evaluate f at all critical numbers in [a,b] AND at the endpoints a and b. The largest value is the absolute maximum, smallest is the absolute minimum.

Concept Tested: Candidates Test

7. A local maximum can occur at x = c if:

  1. f'(c) > 0
  2. f'(c) < 0
  3. f'(c) = 0 or f'(c) does not exist
  4. f''(c) > 0
Show Answer

The correct answer is C. Local extrema can only occur at critical points—where f'(c) = 0 or f'(c) doesn't exist. Not all critical points are extrema, but all local extrema are critical points.

Concept Tested: Local Maximum

8. Find all critical numbers of f(x) = x³ − 3x.

  1. x = 0
  2. x = 1
  3. x = −1 and x = 1
  4. x = 0, x = −1, and x = 1
Show Answer

The correct answer is C. f'(x) = 3x² − 3 = 3(x² − 1) = 3(x−1)(x+1). Setting f'(x) = 0: x = 1 or x = −1.

Concept Tested: Where f' Zero

9. The average rate of change of f(x) = x² from x = 1 to x = 5 is:

  1. 4
  2. 6
  3. 8
  4. 12
Show Answer

The correct answer is B. Average rate = [f(5) − f(1)]/(5 − 1) = (25 − 1)/4 = 24/4 = 6. By MVT, f'(c) = 2c = 6 for some c, giving c = 3.

Concept Tested: Average vs Instant MVT

10. If f is continuous on [0, 4] with f(0) = 2 and f(4) = 10, the MVT guarantees:

  1. f(c) = 6 for some c in (0, 4)
  2. f'(c) = 2 for some c in (0, 4)
  3. f(c) = 0 for some c in (0, 4)
  4. f has a maximum at c = 2
Show Answer

The correct answer is B. MVT: f'(c) = [f(4) − f(0)]/(4 − 0) = (10 − 2)/4 = 2 for some c in (0, 4). The instantaneous rate equals the average rate somewhere.

Concept Tested: MVT Statement