AP Calculus BC 1.16 Working with the Intermediate Value Theorem (IVT) - MCQs - Exam Style Questions
Calc-Ok Question
The function \(f\) is continuous on \([a,b]\) with \(a<b\) and \(f(a)<0<f(b)\). Which statement must be true?
(A) There exists \(x\in(a,b)\) with \(f(x)=0\)
(B) There exists \(x\in(a,b)\) with \(f'(x)=\dfrac{f(b)-f(a)}{b-a}\)
(C) \(f'(x)>0\) for all \(x\in(a,b)\)
(D) \(f(a)<f(x)<f(b)\) for all \(x\in(a,b)\)
(B) There exists \(x\in(a,b)\) with \(f'(x)=\dfrac{f(b)-f(a)}{b-a}\)
(C) \(f'(x)>0\) for all \(x\in(a,b)\)
(D) \(f(a)<f(x)<f(b)\) for all \(x\in(a,b)\)
▶️ Answer/Explanation
Detailed solution
By the Intermediate Value Theorem, continuity with \(f(a)<0<f(b)\) implies some \(c\in(a,b)\) where \(f(c)=0\).
Other options need differentiability or monotonicity not provided.
✅ Answer: (A)
Calc-Ok Question
Selected values of the continuous function \(g\) are given:
| \(x\) | 2 | 3 | 4 |
| \(g(x)\) | \(-2\) | \(1\) | \(-4\) |
Which of the following statements must be true?
(A) The minimum of \(g\) on \([2,4]\) is \(-4\).
(B) \(g'(x)=-1\) has at least one solution in \([2,4]\).
(C) \(g(x)=0\) has at least two solutions in \([2,4]\).
(D) \(g'(x)=0\) has at least two solutions in \([2,4]\).
(B) \(g'(x)=-1\) has at least one solution in \([2,4]\).
(C) \(g(x)=0\) has at least two solutions in \([2,4]\).
(D) \(g'(x)=0\) has at least two solutions in \([2,4]\).
▶️ Answer/Explanation
Detailed solution
By the Intermediate Value Theorem (continuity), a zero exists in \((2,3)\) since \(g(2)<0<g(3)\), and another in \((3,4)\) since \(g(3)>0>g(4)\).
Thus there are at least two zeros in \([2,4]\). The other statements need differentiability or extra information and are not guaranteed.
✅ Answer: (C)