Second Derivative, Concavity, and Inflection Points Practice

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Designed for 12th-grade calculus students. Work through the problems in order to determine second derivatives, intervals of concavity, and coordinates of inflection points.

Find the second derivative of f(x) = 4x^3 - 5x^2 + 7x - 1.
Find all candidate inflection points for f(x) = x^3 - 3x^2 + 4 by solving f''(x) = 0.
For f(x) = x^3 - 3x^2 + 4, determine the intervals where the function is concave down and concave up.
Identify the coordinates of the inflection point for f(x) = x^3 - 3x^2 + 4.
Determine whether f(x) = x^4 has an inflection point at x = 0. Explain why or why not.
Given f'(x) = 0 at x = 3 and f''(3) = -8, classify the point at x = 3 using the second derivative test.

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Second Derivative, Concavity, and Inflection Points Practice

Answer key

Find the second derivative of f(x) = 4x^3 - 5x^2 + 7x - 1.
f''(x) = 24x - 10
Find all candidate inflection points for f(x) = x^3 - 3x^2 + 4 by solving f''(x) = 0.
x = 1
For f(x) = x^3 - 3x^2 + 4, determine the intervals where the function is concave down and concave up.
Concave down on (-infinity, 1); concave up on (1, infinity).
Identify the coordinates of the inflection point for f(x) = x^3 - 3x^2 + 4.
(1, 2)
Determine whether f(x) = x^4 has an inflection point at x = 0. Explain why or why not.
No. While f''(0) = 0, f''(x) = 12x^2 is positive for both x < 0 and x > 0, so concavity does not change.
Given f'(x) = 0 at x = 3 and f''(3) = -8, classify the point at x = 3 using the second derivative test.
Local maximum, because f''(3) < 0 indicates the curve is concave down at that stationary point.