Home / AP Calculus BC : 5.9 Connecting a Function, Its First Derivative, and  Its Second Derivative- Exam Style questions with Answer- MCQ

AP Calculus BC : 5.9 Connecting a Function, Its First Derivative, and  Its Second Derivative- Exam Style questions with Answer- MCQ

AP Calculus BC-Questions and Answers - All Topics

Question

The graph of \( f” \), the second derivative of the function \( f \), is shown above.
Which of the following could be the graph of \( f \)?

A
B
C
D

▶️ Answer/Explanation

Solution

Correct Answer: D

Since the graph of \( f”(x) \) is a parabola opening upwards and has a minimum at the origin, the second derivative is negative before \( x = 0 \) and positive after.

This means that the function \( f \) is concave down when \( x < 0 \) and concave up when \( x > 0 \), with a point of inflection at \( x = 0 \).

Among the choices, only option D shows a function with this concavity behavior. Therefore, the correct answer is D.

Question

The graph of \( y’ = 20x^{3} – 5x^{4} \) has a point of inflection at:

(A) (0, 0) only
(B) (3, 162) only
(C) (4, 256) only
(D) (0, 0) and (3, 162)
(E) (0, 0) and (4, 256)

▶️ Answer/Explanation

Answer: B

Hint:

Given: \( y’ = 20x^3 – 5x^4 \)
Hence \( y” = 60x^2 – 20x^3 = 20x^2(3 – x) \)

Set \( y” = 0 \) and analyze sign changes:
\( y” = 0 \) at \( x = 0 \) and \( x = 3 \)
However, \( y” \) changes sign only at \( x = 3 \), indicating a point of inflection there.

Therefore, the only point of inflection is (3, 162) as \(x=3\)

Question 

If the graph of \(y=x^{3}+ax^{2}+bx-4\) has a point of inflection at (1, -6  ), what is the value of b?
(A) –3                     (B) 0                   (C) 1                       (D) 3                 (E) It cannot be determined from the information given.

▶️ Answer/Explanation
Answer: B
Hint:
The second derivative of the function is:
\( y”(x) = 6x + 2a \)
At a point of inflection, \( y”(x) = 0 \). Substituting \( x = 1 \):
\( 6(1) + 2a = 0 \Rightarrow 2a = -6 \Rightarrow a = -3 \)
Now use the original equation \( y = x^3 + ax^2 + bx – 4 \) and substitute the point \( (1, -6) \):
\( -6 = 1^3 + (-3)(1^2) + b(1) – 4 \Rightarrow -6 = 1 – 3 + b – 4 \)
\( -6 = -6 + b \Rightarrow b = 0 \)

Question

 A force of 10 pounds is required to stretch a spring 4 inches beyond its natural length. Assuming Hooke’s law applies, how much work is done in stretching the spring from its natural length to 6
inches beyond its natural length?

(A) 60.0 inch-pounds
(B) 45.0 inch-pounds
(C) 40.0 inch-pounds
(D) 15.0 inch-pounds
(E) 7.2 inch-pounds

▶️ Answer/Explanation
Solution
First, find the spring constant \( k \) using Hooke’s Law \( F = kx \):
\( 10 = k \times 4 \Rightarrow k = 2.5 \) lb/in
Work done to stretch the spring is given by \( W = \frac{1}{2}kx^2 \):
\( W = \frac{1}{2} \times 2.5 \times 6^2 = \frac{1}{2} \times 2.5 \times 36 = 45 \) inch-pounds
✅ Answer: B) 45.0 inch-pounds
More AP Calculus BC MCQ Questions..
Scroll to Top