Key takeaway: The first derivative test classifies a critical point by watching the sign of the derivative around it: plus to minus gives a local maximum, minus to plus gives a local minimum, no change gives neither.
Grade 12 – AP Calculus ABAP Calculus AB Unit 5Calculus
Definition of First Derivative Test
The first derivative test reads a graph the way a hiker reads a trail. If the slope is positive before a critical point and negative after it, the trail climbs and then descends, so the point is a local peak. If the slope goes from negative to positive, it is a local valley.
The method needs only the sign of the derivative on each side, which is why a sign chart is the standard tool. It works even at points where the second derivative is zero or does not exist.
Formula and notation
Let \(k\) be a critical number of a continuous function \(y=w(x)\). Then
\[ w^{\prime}:\ (+)\to(-)\ \Rightarrow\ \text{local max},\qquad w^{\prime}:\ (-)\to(+)\ \Rightarrow\ \text{local min}. \]
If the sign does not change at \(k\), the point is neither.
Key facts
- Build a sign chart: test one input in each interval between critical numbers.
- A positive derivative means the function rises; a negative derivative means it falls.
- The local extreme value is found by substituting the critical number into the original function.
- Unlike the second derivative method, this test also works where the derivative does not exist.
First Derivative Test example
Use the sign of the derivative to classify the critical points of \(g(x)=2x^3+3x^2-36x+5\).
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Differentiate and factor: \(g^{\prime}(x)=6x^2+6x-36=6(x+3)(x-2)\), giving critical numbers \(-3\) and \(2\).
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Sample one input per interval: \(g^{\prime}(-4)=36\), \(g^{\prime}(0)=-36\), \(g^{\prime}(3)=36\).
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The signs run plus, minus, plus, so \(x=-3\) is a local maximum and \(x=2\) is a local minimum.
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Heights: \(g(-3)=-54+27+108+5=86\) and \(g(2)=16+12-72+5=-39\).
The function peaks locally at 86 when \(x=-3\) and bottoms locally at \(-39\) when \(x=2\).
Common mistake
Watch out
A student finds that \(h(x)=(x-4)^3\) has \(h^{\prime}(4)=0\) and declares a local extreme at \(x=4\).
Why it is wrong
The derivative \(3(x-4)^2\) stays positive on each side of 4, so its sign never switches. The graph flattens briefly and keeps climbing, with no peak or valley. A zero derivative alone is not enough; the first derivative test demands a sign change.
Related terms and lessons
Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus