AP Calculus BC 2.4 Connecting Differentiability and Continuity- Exam Style Questions MCQs - New Syllabus
▶️ Answer/Explanation
Detailed solution
Continuity at the joints:
• At \(x=0\): \(\lim_{x\to 0^-}f(x)=1\), \(\ f(0)=0^2+1=1\), \(\ \lim_{x\to 0^+}f(x)=1\) ⇒ continuous.
• At \(x=1\): \(\lim_{x\to 1^-}f(x)=1^2+1=2\), \(\ f(1)=2\), \(\ \lim_{x\to 1^+}f(x)=1+1=2\) ⇒ continuous.
• Each piece is continuous on its interior ⇒ \(f\) is continuous for all \(x\).
Differentiability at the joints:
• At \(x=0\): left derivative \(0\) (from constant piece), right derivative \(2x|_{x=0}=0\) ⇒ differentiable.
• At \(x=1\): left derivative \(2x|_{x=1}=2\), right derivative \(1\) ⇒ mismatch ⇒ not differentiable.
✅ Correct: B
No-CalcQuestion
The function \[ f(x)= \begin{cases} x^{3}, & x<0,\\[2pt] x^{2}, & 0\le x\le 2,\\[2pt] 4, & x>2 \end{cases} \] is differentiable for which values of \(x\)?
(A) all values of \(x\)
(B) all values of \(x\) except \(x=0\)
(C) all values of \(x\) except \(x=2\)
(D) all values of \(x\) except \(x=0\) and \(x=2\)
(B) all values of \(x\) except \(x=0\)
(C) all values of \(x\) except \(x=2\)
(D) all values of \(x\) except \(x=0\) and \(x=2\)
▶️ Answer/Explanation
Detailed solution
Check potential problem points \(x=0\) and \(x=2\).
At \(x=0\):
• Continuity: \( \lim_{x\to 0^-}x^{3}=0=\lim_{x\to 0^+}x^{2}=f(0) \Rightarrow \) continuous.
• Derivatives: \(f’_-(0)=\lim_{x\to 0^-}3x^{2}=0\), \(f’_+(0)=\lim_{x\to 0^+}2x=0\).
• Left = right ⇒ differentiable at \(x=0\).
At \(x=2\):
• Continuity: \( \lim_{x\to 2^-}x^{2}=4=\lim_{x\to 2^+}4=f(2) \Rightarrow \) continuous.
• Derivatives: \(f’_-(2)=2x\big|_{x=2}=4\), \(f’_+(2)=0\) (since the right piece is constant).
• Left ≠ right ⇒ not differentiable at \(x=2\).
Therefore \(f\) is differentiable for all \(x\) except at \(x=2\).
✅ Answer: (C)