AP®MathsCollege BoardCalculus BCStudy GuidesUnit 5: Analytical Applications of DifferentiationGraphs of Functions & Their DerivativesFirst Derivative Test for Local Extrema

First Derivative Test for Local Extrema (College Board AP® Calculus BC): Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

College Board Calculus BCCollege Board Calculus AB

First derivative test

  • Local extrema (minimums and maximums) are critical points

    • This means the first derivative is equal to zero at these points

  • However, there are some points that have a first derivative of zero but are not local extrema

    • E.g. On the graph of y equals x cubed, the first derivative is zero at x equals 0,

      • but it is not a minimum or maximum, it is a point of inflection

  • You can use the first derivative test to classify critical points

What is the first derivative test?

  • The first derivative test checks the sign of the first derivative just before and just after the critical point

  • If x equals a is a critical point of f open parentheses x close parentheses (i.e. f to the power of apostrophe open parentheses a close parentheses equals 0) and if

    • f to the power of apostrophe open parentheses x close parentheses changes sign from positive to negative at x equals a,

      • then f open parentheses x close parentheses has a local maximum at x equals a

    • f to the power of apostrophe open parentheses x close parentheses changes sign from negative to positive at x equals a,

      • then f open parentheses x close parentheses has a local minimum at x equals a

    • f to the power of apostrophe open parentheses x close parentheses does not change sign at x equals a,

      • then f open parentheses x close parentheses has a point of inflection at x equals a

How do I use the first derivative test to classify critical points?

  • First find the critical points where f to the power of apostrophe open parentheses x close parentheses equals 0

  • Then find the values of the first derivative:

    • at an x value slightly to the left of the critical point

    • at an x value slightly to the right of the critical point

  • If the first derivative changes (from left to right):

    • from positive to negative, it is a local maximum

    • from negative to positive, it is a local minimum

  • If the sign stays the same on both sides of the critical point, it is a point of inflection

f to the power of apostrophe open parentheses x close parentheses before critical point

f to the power of apostrophe open parentheses x close parentheses at critical point

f to the power of apostrophe open parentheses x close parentheses after critical point

Type of critical point

Positive

north east arrow

Zero

long dash

Negative

south east arrow

Maximum

table row blank long dash blank row north east arrow blank south east arrow end table

Negative

south east arrow

Zero

long dash

Positive

north east arrow

Minimum

table row south east arrow blank north east arrow row blank long dash blank end table

Negative

south east arrow

Zero

long dash

Negative

south east arrow

Point of inflection

table row south east arrow blank blank row blank long dash blank row blank blank south east arrow end table

Positive

north east arrow

Zero

long dash

Positive

north east arrow

Point of inflection

table row blank blank north east arrow row blank long dash blank row north east arrow blank blank end table

Examiner Tips and Tricks

If you need to use the first derivative test on a non-calculator question, then choose easy-to-use values to substitute. However, make sure the values are still close enough to the critical point. If they are too far away, then they might cross another critical point, which affects the answer. Select the values so that they are between critical points.

For example, if the critical points are at x equals negative 1 and x equals 1, then when using the first derivative test for x equals negative 1 you need to make sure the value to the right is less than 1. You could pick to use x equals negative 2 and x equals 0 to test x equals negative 1, and x equals 0 and x equals 2 to test x equals 1.

If it is a calculator question, then you can also choose values that are very close to the critical point, such as x equals negative 1.1 and x equals negative 0.9 to test x equals negative 1.

Worked Example

Find the coordinates of the critical points on the graph of f open parentheses x close parentheses equals 2 x cubed plus 3 x squared minus 12 x plus 1, and classify the nature of each point using the first derivative test.

Answer:

Find the derivative of the function

f to the power of apostrophe open parentheses x close parentheses equals 6 x squared plus 6 x minus 12

Find the critical points, where f to the power of apostrophe open parentheses x close parentheses equals 0

table row cell 6 x squared plus 6 x minus 12 end cell equals 0 row cell x squared plus x minus 2 end cell equals 0 row cell open parentheses x plus 2 close parentheses open parentheses x minus 1 close parentheses end cell equals 0 end table

x equals negative 2 and x equals 1

Find the corresponding y values using f open parentheses x close parentheses

f open parentheses negative 2 close parentheses equals 21

f open parentheses 1 close parentheses equals negative 6

Critical points at (-2, 21) and (1, -6)

Classify the points by checking the derivative a little bit to the left and right of each point

Classifying (-2, 21)

f apostrophe open parentheses negative 2.1 close parentheses equals 6 open parentheses negative 2.1 close parentheses squared plus 6 open parentheses negative 2.1 close parentheses minus 12 equals 1.86

f apostrophe open parentheses negative 1.9 close parentheses equals 6 open parentheses negative 1.9 close parentheses squared plus 6 open parentheses negative 1.9 close parentheses minus 12 equals negative 1.74

If you don't have a calculator you could do the same test by considering f to the power of apostrophe open parentheses negative 3 close parentheses and f to the power of apostrophe open parentheses 0 close parentheses

Choose convenient values (like x equals 0); just make sure you don't 'jump across' another critical point!

Classifying (1, -6)

f apostrophe open parentheses 0.9 close parentheses equals negative 1.74

f apostrophe open parentheses 1.1 close parentheses equals 1.86

If you don't have a calculator you could do the same test by considering f to the power of apostrophe open parentheses 0 close parentheses and f to the power of apostrophe open parentheses 2 close parentheses

Summarize your findings

The sign of the first derivative changes from positive to negative around the point (-2, 21); therefore, by the first derivative test, it is a local maximum

The sign of the first derivative changes from negative to positive around the point (1, -6); therefore, by the first derivative test, it is a local minimum

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Author
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Jamie Wood

Author: Jamie Wood

Expertise: Maths Content Creator

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.