Second Derivative Test
In short: Second derivative test classifies a point where f’ equals zero by the sign of f” there: negative means a local maximum, positive means a local minimum, and zero is inconclusive.
AP Calculus ABCalculusAP Calculus AB Unit 5
What is Second Derivative Test?
The second derivative test asks how the curve bends at a flat spot. A flat tangent on a cap shape (bending downward) sits on top of a hill, while a flat tangent on a cup shape (bending upward) sits at the bottom of a valley.
Suppose \(f^{\prime}(c) = 0\) and \(f^{\prime\prime}\) exists near \(c\). If \(f^{\prime\prime}(c) < 0\), then \(f(c)\) is a local maximum. If \(f^{\prime\prime}(c) > 0\), it is a local minimum. If \(f^{\prime\prime}(c) = 0\), this test says nothing and you must switch to another method.
How to spot it
- The problem gives or asks for f” and wants a quick classification of critical points.
- A polynomial has easy second derivative, so plugging numbers beats building a sign chart.
- The question uses the words concave up, concave down and local extremum together.
- Only critical points with a zero first derivative are being tested, not corners or cusps.
The method
- Find \(f^{\prime}(x)\) and solve \(f^{\prime}(x) = 0\) for the critical numbers.
- Differentiate again to get \(f^{\prime\prime}(x)\).
- Evaluate \(f^{\prime\prime}\) at each critical number and read its sign.
- Conclude using the second derivative test; if the value is 0, use the first derivative test instead.
Second Derivative Test: solved example
Problem
Use the second derivative test on \(f(x) = x^4 – 8x^2 + 3\).
Solution
\(f^{\prime}(x) = 4x^3 – 16x = 4x(x – 2)(x + 2)\), so the critical numbers are \(-2\), \(0\), \(2\). Also \(f^{\prime\prime}(x) = 12x^2 – 16\).
\(f^{\prime\prime}(0) = -16 < 0\): local maximum \(f(0) = 3\). \(f^{\prime\prime}(\pm 2) = 48 - 16 = 32 > 0\): local minima, and \(f(\pm 2) = 16 – 32 + 3 = -13\).
Try it
Question 1
For \(f(x) = x^3 – 6x^2\), classify the critical number \(x = 4\).
Show the answer
\(f^{\prime}(x) = 3x^2 – 12x = 0\) at 4, and \(f^{\prime\prime}(x) = 6x – 12\). Since \(f^{\prime\prime}(4) = 12 > 0\), it is a local minimum, \(f(4) = -32\).
Question 2
For \(g(x) = x^4\), what does the test say at \(x = 0\)?
Show the answer
\(g^{\prime\prime}(x) = 12x^2\) gives \(g^{\prime\prime}(0) = 0\), so the test is inconclusive. The first derivative test shows a minimum there.
Second Derivative Test vs. First Derivative Test
This shortcut needs only one number, \(f^{\prime\prime}(c)\), but it gives up when that number is zero and cannot handle points where \(f^{\prime}\) does not exist. The sign-chart method takes longer and always works for a continuous function.
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