AP Calculus AB: 1.16 Working with the Intermediate Value Theorem (IVT)

Question

A polynomial p(x) has a relative maximum at (-2,4), a relative minimum at (1,1), a relative maximum at (5,7) and no other critical points. How many zeros does p(x) have?

A) One
B) Two
C) Three
D) Four
E) Five

▶️ Answer/Explanation

Solution

Correct Answer: B

1. The polynomial must be at least degree 4 (three critical points)

2. Analyzing the behavior:

Starts from ∞ (since it has a max at (-2,4))

Crosses x-axis once to reach min at (1,1)

Crosses x-axis again to reach max at (5,7)

Ends at ∞ (since it’s an even degree polynomial)

3. Therefore, the polynomial must cross the x-axis exactly twice

Question

The function f is continuous on the closed interval [0,2] and has values that are given in the table below. The equation f(x) = 1/2 must have at least two solutions in the interval [0,2] if k =

A) 0
B) 1/2
C) 1
D) 2
E) 3

▶️ Answer/Explanation

Solution

Correct Answer: A

1. The function is continuous on [0,2] with f(0) = 1, f(1) = k, and f(2) = 2

2. For f(x) = 1/2 to have at least two solutions:

– There must be a solution between x=0 and x=1 (requires k < 1/2)

– There must be a solution between x=1 and x=2 (requires k < 1/2)

3. Therefore, k must be less than 1/2

4. Among the options, only A (k=0) satisfies k < 1/2

Question

If a function f is continuous for all x and if f has a relative maximum at (-1, 4) and a relative minimum at (3, -2), which of the following statements must be true?

A) The graph of f has a point of inflection somewhere between x = -1 and x = 3
B) f'(-1) = 0
C) The graph of f has a horizontal asymptote
D) The graph of f has a horizontal tangent line at x = 3
E) The graph of f intersects both axes

▶️ Answer/Explanation

Solution

Correct Answer: E

• A) Not necessarily true (inflection point not guaranteed)

• B) True but not always (f may not be differentiable at x=-1)

• C) No information about asymptotes

• D) True but not always (f may not be differentiable at x=3)

• E) Must be true (continuous function with max at y=4 and min at y=-2 crosses x-axis)

Question

Let g be a continuous function on the closed interval [0,1]. Let g(0)=1 and g(1)=0. Which of the following is NOT necessarily true?

A) There exists a number h in [0,1] such that g(h) ≥ g(x) for all x in [0,1]
B) For all a and b in [0,1], if a=b, then g(a)=g(b)
C) There exists a number h in [0,1] such that g(h)=1/2
D) There exists a number h in [0,1] such that g(h)=3/2
E) For all h in (0,1), \(\lim_{x \to h}g(x) = g(h)\)

▶️ Answer/Explanation

Solution

Correct Answer: D

• A) True (Extreme Value Theorem guarantees absolute maximum)

• B) True (definition of a function)

• C) True (IVT since 1/2 is between g(0)=1 and g(1)=0)

• D) False (3/2 > maximum possible value of g)

• E) True (definition of continuity)

AP Calculus AB: 1.16 Working with the Intermediate Value Theorem (IVT) – Exam Style questions with Answer- MCQ

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