CNothing can be concluded without knowing f explicitly
D|f(5) − f(2)| = 3, because f' must equal 3 at some interior point
By MVT, f(5) − f(2) = f'(c)(5 − 2) for some c ∈ (2, 5). Taking absolute values: |f(5) − f(2)| = |f'(c)| · 3 ≤ 3 · 3 = 9. The bound is an inequality, not an equality (option A), because f' need not be constantly 3. Option D confuses the existence of a point where f' equals the average rate of change with the value of the change itself. This derivative-bounding application is MVT's primary analytical power.
Question 2 Multiple Choice
To derive the MVT from Rolle's Theorem, an auxiliary function g(x) is defined by subtracting the secant line from f. What property of g makes Rolle's Theorem applicable?
Ag is continuous and differentiable, satisfying Rolle's regularity conditions
Bg(a) = g(b) = 0, satisfying Rolle's requirement that the function starts and ends at the same height
Cg is strictly increasing on (a, b), guaranteeing an interior zero of g'
Dg'(x) = f'(x) for all x, so the derivative information is preserved
Rolle's Theorem requires f(a) = f(b). By subtracting the secant line — setting g(x) = f(x) − [f(a) + ((f(b)−f(a))/(b−a))(x−a)] — both g(a) and g(b) equal zero, satisfying this requirement. Option A is true but not novel (differentiability is inherited from f). Option D is wrong: g'(x) = f'(x) minus the secant slope, not equal to f'(x). Rolle's then guarantees g'(c) = 0, which translates back to f'(c) = [f(b)−f(a)]/(b−a).
Question 3 True / False
If f'(x) = 0 for all x in an open interval (a, b), then f is constant on (a, b).
TTrue
FFalse
Answer: True
This follows directly from MVT. For any two points x₁ < x₂ in (a, b), MVT gives f(x₂) − f(x₁) = f'(c)(x₂ − x₁) = 0. So f(x₂) = f(x₁) for any pair of points, meaning f is constant. This is the rigorous justification for the calculus claim that antiderivatives of the same function differ only by a constant — a fact that undergirds the entire theory of definite integration.
Question 4 True / False
The Mean Value Theorem guarantees that the derivative equals the average rate of change at exactly one interior point.
TTrue
FFalse
Answer: False
MVT guarantees existence of *at least one* such point c, not exactly one. A function could satisfy f'(c) = [f(b)−f(a)]/(b−a) at multiple interior points — for example, a horizontal line (f constant) satisfies f' = 0 = average rate of change everywhere on the interval. The theorem is an existence result, not a uniqueness result. Mistaking 'at least one' for 'exactly one' is a common misreading.
Question 5 Short Answer
Explain why the Mean Value Theorem is more than a geometric observation and what analytical work it actually does.
Think about your answer, then reveal below.
Model answer: MVT converts local information (derivative values at individual points) into global information about the function's behavior over an interval. The geometric picture shows the result is plausible, but the analytical power is the inequality |f(b)−f(a)| ≤ M|b−a|: a function cannot change faster than its maximum derivative rate. This enables bounding functions in error analysis, proving monotonicity, and establishing uniqueness of antiderivatives.
The kinematic analogy (you were exactly at the average speed at some instant) makes MVT vivid but doesn't convey its usefulness. The analytical leverage comes from being able to *bound* how much a function changes given a bound on its derivative. This appears throughout analysis: proving convergence, bounding approximation errors, and establishing monotonicity all reduce to MVT-style arguments where derivative bounds translate into function-value bounds.