AP Calculus AB 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist - MCQs - Exam Style Questions
No-Calc Question
The piecewise function \(f\) is defined as shown. At what values of \(x\) is \(f\) not differentiable?
\[ f(x)= \begin{cases} x^{2}+3x+2, & x<-1,\\ 1-|x|, & -1\le x\le 1,\\ x^{2}-1, & x>1. \end{cases} \]
(A) \(0\) only
(B) \(1\) only
(C) \(0\) and \(1\) only
(D) \(-1, 0,\) and \(1\)
▶️ Answer/Explanation
Check junctions \(-1,0,1\).
At \(-1\): \(f\) is continuous and \(f’_{-}(-1)=2(-1)+3=1,\ f’_{+}(-1)=1\) ⇒ differentiable.
At \(0\): left slope \(1\), right slope \(-1\) ⇒ corner ⇒ not differentiable.
At \(1\): left slope \(-1\), right slope \(2(1)=2\) ⇒ kink ⇒ not differentiable.
✅ Answer: (C)
No-Calc Question
Let \(f\) be the function defined below.
\(f(x)=\begin{cases} -x^{2}+3,& x\le 5\\ -10x+28,& x>5 \end{cases}\)
Which of the following statements about \(f\) is true?
(A) \(f\) is continuous and differentiable at \(x=5\).
(B) \(f\) is continuous but not differentiable at \(x=5\).
(C) \(f\) is differentiable but not continuous at \(x=5\).
(D) \(f\) is defined but neither continuous nor differentiable at \(x=5\).
(A) \(f\) is continuous and differentiable at \(x=5\).
(B) \(f\) is continuous but not differentiable at \(x=5\).
(C) \(f\) is differentiable but not continuous at \(x=5\).
(D) \(f\) is defined but neither continuous nor differentiable at \(x=5\).
▶️ Answer/Explanation
Continuity at \(5\):
Left value \(f(5^{-})=-25+3=-22\). Right value \(f(5^{+})=-50+28=-22\). So continuous.
Derivatives: \((-x^{2}+3)’\!=\!-2x\Rightarrow -10\) at \(5\).
\((-10x+28)’\!=\!-10\) at \(5\). Derivatives match → differentiable.
✅ Answer: (A)