Second derivative; max/min test practice questions.

For f(x) = x³ − 6x² + 5, determine the second derivative f''(x).

• 6x − 12
• 3x² − 12x
• 6x² − 12
• 6x − 6

Answer: 6x − 12

f'(x) = 3x² − 12x, so f''(x) = 6x − 12.

At a stationary point of f, if f''(x) > 0, determine the nature of that point.

• Local maximum
• Local minimum
• Point of inflexion
• Cannot be determined

Answer: Local minimum

A positive second derivative means the curve is concave up, so the stationary point is a local minimum.

For f(x) = 2x³ − 9x² + 12x, the stationary points occur at x = 1 and x = 2. Using f''(x) = 12x − 18, determine the nature of the stationary point at x = 1.

• Local minimum
• Local maximum
• Point of inflexion
• Neither

Answer: Local maximum

f''(1) = 12 − 18 = −6 < 0, so the curve is concave down, giving a local maximum.

The second derivative of a function is f''(x) = 6x. Determine the x-coordinate of the possible point of inflexion.

• x = 6
• x = 1
• x = 0
• x = −6

Answer: x = 0

A point of inflexion requires f''(x) = 0, so 6x = 0 gives x = 0 (with a sign change confirmed).

For f(x) = e^x, determine the second derivative f''(x).

• e^x
• x·e^(x−1)
• 1
• e^(2x)

Answer: e^x

Every derivative of e^x is e^x, so f''(x) = e^x.

For f(x) = x⁴, at the stationary point x = 0 the second derivative test gives f''(0) = 0. Determine the correct conclusion.

• It must be a maximum
• It must be a point of inflexion
• Inconclusive; a minimum
• It cannot be a stationary point

Answer: Inconclusive; a minimum

f''(0) = 0 makes the test inconclusive; examining f (or the sign of f') shows x = 0 is actually a minimum.