Home / AP Calculus AB 1.16 Working with the Intermediate Value Theorem (IVT) – MCQs

AP Calculus AB 1.16 Working with the Intermediate Value Theorem (IVT) - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

\(x\)\(f(x)\)
\(-5\)\(-9\)
\(0\)\(1\)
\(2\)\(5\)

The table above gives values of a continuous function \(f\) at selected values of \(x\). Based on the information in the table, which of the following statements must be true?

(A) \(f\) has at most one zero.
(B) \(f\) has a relative maximum at \(x=2\).
(C) There exists a value \(c\), where \(-5<c<2\), such that \(f(c)=4\).
(D) There exists a value \(c\), where \(-5<c<2\), such that \(f'(c)=2\).

▶️ Answer/Explanation
Since \(f\) is continuous on \([-5,2]\) and \(f(-5)=-9<4<f(2)=5\), the Intermediate Value Theorem guarantees a \(c\in(-5,2)\) with \(f(c)=4\).
The other statements are not forced by the data provided.
✅ Answer: (C)

Calc-OkQuestion

Let \(f\) be a function such that \(f(1)=-2\) and \(f(5)=7\). Which of the following conditions ensures that \(f(c)=0\) for some \(c\in(1,5)\)?

(A) \(\displaystyle \int_{1}^{5} f(x)\,dx\) exists
(B) \(f\) is increasing on \([1,5]\)
(C) \(f\) is continuous on \([1,5]\)
(D) \(f\) is defined for all \(x\in[1,5]\)

▶️ Answer/Explanation

With \(f(1)=-2<0<7=f(5)\), continuity on \([1,5]\) guarantees a zero by the Intermediate Value Theorem. The other conditions alone do not ensure a sign change is attained.
✅ Answer: (C)

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