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

First Derivative Test for Local Extrema (College Board AP® Calculus AB)

Revision Note

Jamie Wood

Expertise

Maths

First derivative test

  • We know that 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 which 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

  • This means a better definition for local minimums and maximums is:

  • 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

What is the first derivative test?

  • Using the relationship between local extrema and the first derivative described above, we can find local minimums and maximums

  • 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

/

Zero

⁻

Negative

\

Maximum

/—\

Negative

\

Zero

_

Positive

/

Minimum

\_/

Negative

\

Zero

_

Negative

\

Point of inflection

backslash long dash space space space space space space space backslash

Positive

/

Zero

_

Positive

/

Point of inflection

space space space space space space space space divided by divided by to the power of stack space space space space space space with bar on top end exponent

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

Changes from positive to negative, so (-2, 21) is a maximum

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

Changes from negative to positive, so (1, -6) is a minimum

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

Local maximum at (-2, 21)

Local minimum at (1, -6)

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Next topic

    Did this page help you?

    Jamie Wood

    Author: Jamie Wood

    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.