Home / AP Calculus BC : 5.6 Determining Concavity of Functions over Their  Domains- Exam Style questions with Answer- MCQ

AP Calculus BC : 5.6 Determining Concavity of Functions over Their  Domains- Exam Style questions with Answer- MCQ

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Question

The first derivative of the function f is defined by \({f}'(x) = (x^2 +1)sin(3x – 1)\) for \(-1.5 < x < 1.5\). On which of the following intervals is the graph of f concave up?

A \((−1.5, −1.341)\) and \((−0.240, 0.964)\)

B \((−1.341, −0.240)\) and \((0.964, 1.5)\)

C \((−0.714, 0.333)\) and \((1.381, 1.5)\)

D \((−1.5, −0.714) and (0.333, 1.381)\)

▶️ Answer/Explanation
Solution
Compute \( f”(x) \): \( f'(x) = (x^2 + 1) \sin(3x – 1) \).
\( f”(x) = 2x \sin(3x – 1) + 3(x^2 + 1) \cos(3x – 1) \).
Test intervals for \( f”(x) > 0 \):
\( x = -1.4205 \) in \( (-1.5, -1.341) \): \( f”(-1.4205) \approx 2.597 > 0 \).
\( x = 0 \) in \( (-0.240, 0.964) \): \( f”(0) \approx 1.62 > 0 \).
\( x = -1 \) in \( (-1.341, -0.240) \): \( f”(-1) \approx -5.55 < 0 \).
\( x = 1 \) in \( (0.964, 1.5) \): \( f”(1) \approx -0.678 < 0 \).
Concave up where \( f”(x) > 0 \): \( (-1.5, -1.341) \) and \( (-0.240, 0.964) \).
✅ Answer: A)

Question

The graph of \( y = f(x) \) on the closed interval [2, 7] is shown below.

Graph of function f(x)

How many points of inflection does this graph have on this interval?

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

▶️ Answer/Explanation

Solution

Correct Answer: C

To find inflection points, we look for where the graph changes concavity.
Starting at \( x = 2 \), the graph is concave up,
then transitions to concave down near a local maximum — that’s the first inflection point.
As the graph continues, it becomes concave up again near a local minimum — that’s the second inflection point.
Finally, the graph becomes concave down again as it descends — that’s the third inflection point.

Therefore, the graph has three points of inflection.

Question

Let f be the function given by \( f(x) = \cos(2x) + \ln(3x) \). What is the least value of x at which the graph of f changes concavity?

A) 0.56
B) 0.93
C) 1.18
D) 2.38
E) 2.44

▶️ Answer/Explanation
Solution

Step 1: Find second derivative
First derivative: \( f'(x) = -2\sin(2x) + \frac{1}{x} \)
Second derivative: \( f”(x) = -4\cos(2x) – \frac{1}{x^2} \)

Step 2: Find where concavity changes
Solve \( f”(x) = 0 \): \( -4\cos(2x) – \frac{1}{x^2} = 0 \)
Numerically solving gives the smallest positive solution at \( x \approx 0.93 \)

Step 3: Verify concavity change
Checking values around 0.93:
For \( x = 0.9 \): \( f”(x) > 0 \) (concave up)
For \( x = 1.0 \): \( f”(x) < 0 \) (concave down)

Conclusion:
The graph changes from concave up to concave down at \( x \approx 0.93 \)

✅ Answer: B) 0.93

Question

 At what value of x does the function \(y=1.2x^{2}-e^{.4x}\)  change concavity?
(A) x = 2.5 ln 6
(B) x = 6 ln 2.4
(C) x = 2.4 ln 12
(D) x = 2.5 ln 15

▶️ Answer/Explanation
Solution

First derivative: \( y’ = 2.4x – 0.4 e^{0.4x} \).

Second derivative: \( y” = 2.4 – 0.16 e^{0.4x} \).

Set \( y” = 0 \): \( 0.16 e^{0.4x} = 2.4 \), so \( e^{0.4x} = 15 \).

Thus: \( 0.4x = \ln 15 \), and \( x = 2.5 \ln 15 \).

Concavity changes as: \( y” > 0 \) for \( x < 2.5 \ln 15 \), \( y” < 0 \) for \( x > 2.5 \ln 15 \).

✅ Answer: D) \( x = 2.5 \ln 15 \)

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