AP Calculus AB: Concavity and Second Derivative Test (Drill 1)

Question 1. Let \( f(x) = x^4 - 4x^3 \). On which of the following intervals is the graph of \( f \) concave up?

• A) \( x > 0 \) only
• B) \( x 2 \) ✓
• C) \( x > 2 \) only
• D) \( 0 < x < 2 \)

Explanation: Choice B is correct. \( f'(x) = 4x^3 - 12x^2 \) and \( f''(x) = 12x^2 - 24x = 12x(x-2) \). The graph is concave up where \( f''(x) > 0 \). Setting \( f''(x) = 0 \) gives \( x = 0 \) and \( x = 2 \). On \( (-\infty, 0) \), both factors \( 12x \) and \( (x-2) \) are negative, so \( f''(x) > 0 \) (concave up). On \( (0, 2) \), \( 12x > 0 \) but \( (x-2) < 0 \), so \( f''(x) 0 \) (concave up). Choice A is incorrect because \( (0, 2) \) is concave down. Choice C is incorrect because it omits the concave-up region on \( (-\infty, 0) \). Choice D is incorrect because \( (0, 2) \) is where the function is concave down.

Question 2. Let \( f(x) = x^3 - 3x^2 + 4 \). At which value of \( x \) does \( f \) have a point of inflection?

• A) \( x = 0 \)
• B) \( x = 1 \) ✓
• C) \( x = 2 \)
• D) \( x = 3 \)

Explanation: Choice B is correct. \( f''(x) = 6x - 6 = 6(x-1) \). Setting \( f''(x) = 0 \) gives \( x = 1 \). Since \( f''(x) < 0 \) for \( x 0 \) for \( x > 1 \), concavity changes at \( x = 1 \), confirming a point of inflection. Choice A is incorrect because \( f''(0) = -6 \neq 0 \); since \( f''(0) \) is not zero and \( f'' \) is not undefined there, \( x = 0 \) cannot be a point of inflection. Choice C is incorrect because \( f''(2) = 6 \neq 0 \); for the same reason, \( x = 2 \) cannot be a point of inflection. Choice D is incorrect because \( f''(3) = 12 \neq 0 \), so \( x = 3 \) cannot be a point of inflection either.

Question 3. Let \( f(x) = x^4 - 8x^2 \). Which of the following correctly classifies the critical point at \( x = 2 \)?

• A) Relative maximum, because \( f''(2) < 0 \)
• B) Relative minimum, because \( f''(2) > 0 \) ✓
• C) Relative maximum, because \( f'(2) = 0 \)
• D) The Second Derivative Test is inconclusive at \( x = 2 \) because \( f''(2) = 0 \)

Explanation: Choice B is correct. \( f'(x) = 4x^3 - 16x = 4x(x-2)(x+2) \), confirming \( x = 2 \) is a critical point. \( f''(x) = 12x^2 - 16 \), so \( f''(2) = 48 - 16 = 32 > 0 \). Since \( f''(2) > 0 \), the graph is concave up at \( x = 2 \), indicating a relative minimum. Choice A is incorrect because \( f''(2) = 32 > 0 \), not negative; a relative maximum requires \( f'' < 0 \) at the critical point. Choice C is incorrect because \( f'(2) = 0 \) identifies \( x = 2 \) as a critical point but does not classify it, the second derivative value is needed. Choice D is incorrect because \( f''(2) = 32 \neq 0 \); the test is inconclusive only when \( f'' = 0 \) at the critical point.

Question 4. For a twice-differentiable function \( g \), it is known that \( g''(4) = 0 \). Which of the following must be true?

• A) \( g \) has a point of inflection at \( x = 4 \).
• B) \( g \) has a relative extremum at \( x = 4 \).
• C) \( x = 4 \) must be a critical point of \( g \).
• D) None of the above must be true. ✓

Explanation: Choice D is correct. \( g''(4) = 0 \) is necessary but not sufficient for any of the conclusions in choices A, B, or C. For a point of inflection, concavity must actually change sign at \( x = 4 \), \( g''(4) = 0 \) alone does not guarantee this. The function \( g(x) = (x-4)^4 \) illustrates this: \( g''(x) = 12(x-4)^2 \), so \( g''(4) = 0 \), yet \( g''(x) \geq 0 \) on both sides of \( x = 4 \) and concavity never changes, so there is no inflection point. Choice A is incorrect because \( g''(4) = 0 \) does not guarantee a sign change in \( g'' \) at \( x = 4 \). Choice B is incorrect because a relative extremum requires \( g'(4) = 0 \) (or undefined) combined with a sign change in \( g' \); \( g''(4) = 0 \) implies nothing about \( g'(4) \). Choice C is incorrect because critical points are defined by \( g'(4) = 0 \) or \( g'(4) \) undefined, knowing \( g''(4) = 0 \) says nothing about the value of \( g'(4) \).

Question 5. A particle moves along a straight line with velocity \( v(t) \) for \( t \geq 0 \). It is given that \( v(t) > 0 \), \( v'(t) > 0 \), and \( v''(t) < 0 \) for all \( t \) in the interval \( (0, 5) \). Which of the following correctly describes the particle’s motion on \( (0, 5) \)?

• A) The particle is moving in the positive direction and slowing down.
• B) The particle is moving in the positive direction, speeding up, and its acceleration is increasing.
• C) The particle is moving in the positive direction, speeding up, and its acceleration is decreasing. ✓
• D) The particle is decelerating.

Explanation: Choice C is correct. \( v(t) > 0 \) means the particle moves in the positive direction. \( v'(t) > 0 \) means velocity is increasing, so the particle is speeding up. \( v''(t) 0 \) means velocity is increasing, so the particle is speeding up, not slowing down. Choice B is incorrect because \( v''(t) < 0 \) means acceleration is decreasing, not increasing. Choice D is incorrect because decelerating means the particle is slowing down, which would require \( v'(t) 0 \), so the particle is still accelerating, just at a diminishing rate.