Key takeaway: The second derivative test classifies a critical point where the first derivative is zero: a negative second derivative means a local maximum, and a positive one means a local minimum.
Grade 12 – AP Calculus ABAP Calculus AB Unit 5Calculus
Definition of Second Derivative Test
The second derivative test classifies a flat spot of a graph by its bend. At a critical number where the tangent is horizontal, a curve that opens downward like a cap peaks there, and a curve that opens upward like a cup bottoms out there.
The second derivative measures the bend. Its sign at the critical number therefore decides the type of extreme, often faster than building a sign chart. The shortcut has a limit: when the second derivative is zero, the test says nothing.
Formula and notation
Suppose \(v^{\prime}(k)=0\). Then
\[ v^{\prime\prime}(k)\lt 0\Rightarrow\text{local max},\qquad v^{\prime\prime}(k)\gt 0\Rightarrow\text{local min},\qquad v^{\prime\prime}(k)=0\Rightarrow\text{no conclusion}. \]
Here \(v^{\prime\prime}\) is the derivative of the derivative.
Key facts
- Start by solving for the critical numbers where the first derivative is zero.
- Cap shape, with a negative second derivative, means a peak; cup shape means a valley.
- If the second derivative is zero, switch to the first derivative method.
- The test applies only where the first derivative equals zero, not where it is undefined.
Second Derivative Test example
Classify the critical points of \(g(x)=x^3-9x^2+15x+2\) with the second derivative.
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Slope formula: \(g^{\prime}(x)=3x^2-18x+15=3(x-1)(x-5)\), which vanishes at 1 and 5.
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Bend formula: \(g^{\prime\prime}(x)=6x-18\).
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At \(x=1\): \(g^{\prime\prime}(1)=-12\), negative. At \(x=5\): \(g^{\prime\prime}(5)=12\), positive.
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Heights: \(g(1)=1-9+15+2=9\) and \(g(5)=125-225+75+2=-23\).
The function has a local maximum of 9 at \(x=1\) and a local minimum of \(-23\) at \(x=5\).
Common mistake
Watch out
For \(h(x)=x^4\), a student finds \(h^{\prime}(0)=0\) and \(h^{\prime\prime}(0)=0\), then claims that \(x=0\) is not an extreme point.
Why it is wrong
A zero second derivative means the second derivative test is inconclusive, not that there is no extreme. The first derivative \(4x^3\) changes from negative to positive at 0, so \(x=0\) is in fact a local minimum. Use the other method when the test gives no answer.
Related terms and lessons
Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus