AP Calculus AB Question of the Day

1. Yes, because $f(0)=f(2)$, so the derivative must be zero everywhere.
2. No, because MVT only applies when $f(0)\ne f(2)$.
3. Yes, because $f$ is continuous on $[0,2]$, differentiable on $(0,2)$, and $\frac{f(2)-f(0)}{2-0}=0$. (correct answer)
4. Yes, because $0$ is between $f(0)$ and $f(2)$.
5. No, because $f'(c)=0$ requires $f$ to have a maximum or minimum.

Explanation: This is a special case of MVT known as Rolle's Theorem, which applies when f(a) = f(b). Here, f is continuous on [0,2] and differentiable on (0,2), with f(0) = f(2) = 1. The average rate of change is (f(2)-f(0))/(2-0) = (1-1)/2 = 0/2 = 0. Since MVT guarantees a point where the derivative equals this average rate, there must exist at least one c in (0,2) where f'(c) = 0. This doesn't mean the derivative is zero everywhere (choice A's error), just at least at one point. A common misconception is that f'(c) = 0 requires a maximum or minimum, but it could also occur at an inflection point. When endpoints have equal function values, MVT simplifies to guaranteeing a horizontal tangent somewhere.