Second derivative; concavity practice questions.

Consider f(x) = 2x³ − 9x² + 12x. Determine the interval on which the graph of f is concave up.

• x < 3/2
• x > 3
• x > 3/2
• x < 3

Answer: x > 3/2

f″(x) = 12x − 18, which is positive when x > 3/2. Concave up means f″(x) > 0.

Let g(x) = x³ + 6x². Calculate g″(−1).

• 6
• −6
• 18
• 9

Answer: 6

g′(x) = 3x² + 12x, so g″(x) = 6x + 12. Substituting x = −1 gives 6(−1) + 12 = 6.

On an interval where f″(x) < 0 for all x, which statement must be true?

• f is decreasing on the interval
• f′ is decreasing on the interval
• f has a local maximum in the interval
• f′(x) < 0 on the interval

Answer: f′ is decreasing on the interval

f″ is the derivative of f′, so f″ < 0 means f′ is decreasing (the graph is concave down). It says nothing about the sign of f or f′ themselves.

Determine the concavity of f(x) = sin x on the interval 0 < x < π.

• Concave up on the whole interval
• Concave up on (0, π/2), concave down on (π/2, π)
• Concave down on (0, π/2), concave up on (π/2, π)
• Concave down on the whole interval

Answer: Concave down on the whole interval

f″(x) = −sin x, and sin x > 0 for 0 < x < π, so f″(x) < 0 throughout: the graph is concave down on the entire interval.

Which of the following functions is concave up for every real value of x?

• f(x) = eˣ
• f(x) = x³
• f(x) = ln x
• f(x) = sin x

Answer: f(x) = eˣ

For f(x) = eˣ, f″(x) = eˣ > 0 for all x, so it is concave up everywhere. x³ changes concavity at 0, ln x is not defined for all x, and sin x alternates concavity.

Consider f(x) = x⁴ − 4x³. Determine the interval on which the graph of f is concave down.

• (−∞, 0)
• (2, ∞)
• (0, 2)
• (0, 4)

Answer: (0, 2)

f″(x) = 12x² − 24x = 12x(x − 2), which is negative exactly when 0 < x < 2.