AP Calculus BC 5.1 Using the Mean Value Theorem - MCQs - Exam Style Questions
No-Calc Question
The function \(f\) is continuous on \([0,5]\) and differentiable on \((0,5)\). Selected values are in the table, and \(c=4\) satisfies the conclusion of the Mean Value Theorem for \(f\) on \([0,5]\). What is the slope of the line tangent to the graph of \(f\) at \(x=4\)?
| \(x\) | 0 | 2 | 3 | 5 |
| \(f(x)\) | 2 | 3 | 6 | 12 |
(B) \(2.8\)
(C) \(4.5\)
(D) \(10\)
▶️ Answer/Explanation
By the Mean Value Theorem, there exists \(c\in(0,5)\) with \(f'(c)=\dfrac{f(5)-f(0)}{5-0}\). Given \(c=4\), this equals the slope at \(x=4\).
\(\displaystyle f'(4)=\dfrac{12-2}{5}=2.\)
✅ Answer: (A)
▶️ Answer/Explanation
By the Mean Value Theorem (function is continuous and differentiable), on any interval \([a,b]\) there exists \(c\in(a,b)\) with \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] Compute average slopes on each listed interval:
\((0,4):\ \dfrac{0-8}{4-0}=-2\) (not \(2\)).
\((4,8):\ \dfrac{2-0}{8-4}=\dfrac{2}{4}=0.5\) (not \(2\)).
\((8,12):\ \dfrac{10-2}{12-8}=\dfrac{8}{4}=2\) ✅ gives \(2\).
\((12,16):\ \dfrac{1-10}{16-12}=\dfrac{-9}{4}\) (not \(2\)).
Hence there must be some \(c\in(8,12)\) with \(f'(c)=2\).
✅ Answer: (C)