According to the Mean Value Theorem, there exists c in (0,4) with f'(c) equal to the average rate of change of f on [0,4]. If f(0) = 2 and f(4) = 18, what is f'(c)?

According to the Mean Value Theorem, there exists c in (0,4) with f'(c) equal to the average rate of change of f on [0,4]. If f(0) = 2 and f(4) = 18, what is f'(c)?

• A. There exists c in (0,4) with f'(c) = 4
• B. There exists c in (0,4) with f''(c) = 4
• C. f'(0) = 4
• D. f'(4) = 4

Answer: (A)

The Mean Value Theorem says that for a differentiable function on [a,b], there is a point c in (a,b) where the tangent slope f'(c) equals the slope of the secant line connecting (a,f(a)) and (b,f(b)). Here, a = 0 and b = 4 with f(0) = 2 and f(4) = 18. The average rate of change is (18 − 2)/(4 − 0) = 16/4 = 4, so there exists c in (0,4) such that f'(c) = 4. This matches the statement that there is a point in (0,4) where the derivative equals 4. The other possibilities aren’t guaranteed by this theorem: it concerns the first derivative at some interior point, not the second derivative or the derivative at endpoints.

The Mean Value Theorem says that for a differentiable function on [a,b], there is a point c in (a,b) where the tangent slope f'(c) equals the slope of the secant line connecting (a,f(a)) and (b,f(b)). Here, a = 0 and b = 4 with f(0) = 2 and f(4) = 18. The average rate of change is (18 − 2)/(4 − 0) = 16/4 = 4, so there exists c in (0,4) such that f'(c) = 4. This matches the statement that there is a point in (0,4) where the derivative equals 4. The other possibilities aren’t guaranteed by this theorem: it concerns the first derivative at some interior point, not the second derivative or the derivative at endpoints.