AP Calculus BC 5.6 Determining Concavity of Functions over Their Domains - Exam Style Questions - MCQs - New Syllabus
Question
Let \(f\) be a function with second derivative \(f”(x)=(x-1)^2(x-2)(x-3)^3\). How many points of inflection does the graph of \(f\) have?
A. Two
B. Three
C. Five
D. Six
B. Three
C. Five
D. Six
▶️ Answer/Explanation
Detailed solution
Inflection points occur where \(f”\) changes sign.
Zeros of \(f”\): \(x=1\) (mult. 2, even ⇒ no sign change), \(x=2\) (mult. 1, odd ⇒ sign change), \(x=3\) (mult. 3, odd ⇒ sign change).
Hence there are \( \boxed{2} \) points of inflection (at \(x=2\) and \(x=3\)).
✅ Correct: A
Question
The function \( f \) has derivative \( f'(x)=x^{2}\tan(1-2x) \) for \( -0.25<x<0.75 \). At which value(s) of \(x\) does the graph of \(f\) have a point of inflection?
(A) \(0.324\) only
(B) \(0.5\) only
(C) \(0\) and \(0.324\)
(D) \(0\) and \(0.5\)
(B) \(0.5\) only
(C) \(0\) and \(0.324\)
(D) \(0\) and \(0.5\)
▶️ Answer/Explanation
Correct answer: (C)
\[ f”(x)=2x\tan(1-2x)-2x^{2}\sec^{2}(1-2x). \] Inflection where \(f”\) changes sign: one root at \(x=0\). For \(x\ne0\), solve \[ \tan(1-2x)=x\sec^{2}(1-2x) \;\Longleftrightarrow\; \sin(2-4x)=2x. \] On \((-0.25,0.75)\) this gives \(x\approx 0.324\). Thus \(x=0\) and \(x\approx 0.324\).