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Graphs of Functions & Their Derivatives

Sarah Miller

Sarah Miller

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Next Topic - Candidates Test for Global Extrema

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Study Guide Overview

This study guide covers the First Derivative Test for finding and classifying local extrema. It explains the relationship between the first derivative and local extrema (maxima and minima), including the concept of critical points. The guide outlines the steps of the First Derivative Test, provides a worked example, practice questions, and a glossary of key terms like local extrema, critical points, and the First Derivative Test itself.

#First Derivative Test

#Table of Contents

  1. Introduction
  2. Relationship Between First Derivative and Local Extrema
  3. First Derivative Test
  4. Worked Example
  5. Practice Questions
  6. Glossary
  7. Summary and Key Takeaways

#Introduction

Understanding how the first derivative of a function is related to its local extrema (maximums and minimums) is crucial in calculus. This study note will explore the first derivative test, a method used to identify and classify these extrema.

#Relationship Between First Derivative and Local Extrema

Key Concept

#Key Concepts

  • Local Extrema: Points where a function reaches a local maximum or minimum.
  • Critical Points: Points where the first derivative of a function is zero or undefined.

#Explanation

Local extrema are found at critical points where the first derivative, f′(x)f'(x)f′(x), is equal to zero. However, not all points where f′(x)=0f'(x) = 0f′(x)=0 are local extrema. For example, the graph of y=x3y = x^3y=x3 has a derivative of zero at x=0x = 0x=0, but this point is not a minimum or maximum; it is a point of inflection.

A more precise definition for local minimums and maximums is:

  • If x=ax = ax=a is a critical point of f(x)f(x)f(x) (i.e., f′(a)=0f'(a) = 0f′(a)=0) and if:
    • f′(x)f'(x)f′(x) changes from positive to negative at x=ax = ax=a, then x=ax = ax=a is a local maximum.
    • f′(x)f'(x)f′(x) changes from negative to positive at x=ax = ax=a, then x=ax = ax=a is a local minimum.
    • If the sign of f′(x)f'(x)f′(x) does not change, x=ax = ax=a is a point of inflection.

#First Derivative Test

#Steps

  1. Find the Critical Points: Solve f′(x)=0f'(x) = 0f′(x)=0 to identify critical points.
  2. Evaluate the First Derivative Around Critical Points:
    • Check the value of f′(x)f'(x)f′(x) slightly to the left of the critical point.
    • Check the value of f′(x)f'(x)f′(x) slightly to the right of the critical point.
  3. Classify the Critical Points:
    • If f′(x)f'(x)f′(x) changes from positive to negative, it is a local maximum.
    • If f′(x)f'(x)f′(x) changes from negative to positive, it is a local minimum.
    • If f′(x)f'(x)f′(x) does not change sign, it is a point of inflection.

#Summary Table

f′(x)f'(x)f′(x) before critical pointf′(x)f'(x)f′(x) at critical pointf′(x)f'(x)f′(x) after critical pointType of Critical Point
PositiveZeroNegativeLocal Maximum
NegativeZeroPositiveLocal Minimum
NegativeZeroNegativePoint of Inflection
PositiveZeroPositivePoint of Inflection

#Worked Example

#Problem

Find the coordinates of the critical points on the graph of f(x)=2x3+3x2−12x+1f(x) = 2x^3 + 3x^2 - 12x + 1f(x)=2x3+3x2−12x+1, and classify the nature of each point using the first derivative test.

#Solution

  1. Find the Derivative: f′(x)=6x2+6x−12f'(x) = 6x^2 + 6x - 12f′(x)=6x2+6x−12

  2. Find the Critical Points: 6x2+6x−12=0x2+x−2=0(x+2)(x−1)=0\begin{aligned} 6x^2 + 6x - 12 &= 0 \\ x^2 + x - 2 &= 0 \\ (x + 2)(x - 1) &= 0 \\ \end{aligned}6x2+6x−12x2+x−2(x+2)(x−1)​=0=0=0​ Thus, x=−2x = -2x=−2 and x=1x = 1x=1.

  3. Evaluate f(x)f(x)f(x) at Critical Points: f(−2)=2(−2)3+3(−2)2−12(−2)+1=21f(1)=2(1)3+3(1)2−12(1)+1=−6\begin{aligned} f(-2) &= 2(-2)^3 + 3(-2)^2 - 12(-2) + 1 = 21 \\ f(1) &= 2(1)^3 + 3(1)^2 - 12(1) + 1 = -6 \\ \end{aligned}f(−2)f(1)​=2(−2)3+3(−2)2−12(−2)+1=21=2(1)3+3(1)2−12(1)+1=−6​ Critical points are (−2,21)(-2, 21)(−2,21) and (1,−6)(1, -6)(1,−6).

  4. Classify the Points:

    • At x=−2x = -2x=−2: f′(−2.1)>0(positive)f′(−1.9)<0(negative)\begin{aligned} f'(-2.1) &> 0 \quad (\text{positive}) \\ f'(-1.9) &< 0 \quad (\text{negative}) \\ \end{aligned}f′(−2.1)f′(−1.9)​>0(positive)<0(negative)​ Changes from positive to negative, so (−2,21)(-2, 21)(−2,21) is a local maximum.

    • At x=1x = 1x=1: f′(0.9)<0(negative)f′(1.1)>0(positive)\begin{aligned} f'(0.9) &< 0 \quad (\text{negative}) \\ f'(1.1) &> 0 \quad (\text{positive}) \\ \end{aligned}f′(0.9)f′(1.1)​<0(negative)>0(positive)​ Changes from negative to positive, so (1,−6)(1, -6)(1,−6) is a local minimum.

#Conclusion

  • Local maximum at (−2,21)(-2, 21)(−2,21).
  • Local minimum at (1,−6)(1, -6)(1,−6).

#Practice Questions

Practice Question
  1. Find the critical points and classify them for the function f(x)=x3−3x2+4f(x) = x^3 - 3x^2 + 4f(x)=x3−3x2+4.
  2. Given f(x)=x4−4x3+6x2f(x) = x^4 - 4x^3 + 6x^2f(x)=x4−4x3+6x2, use the first derivative test to find and classify all local extrema.
  3. For f(x)=ex−x2f(x) = e^x - x^2f(x)=ex−x2, determine the critical points and their nature.

#Glossary

  1. Local Extrema: Points where a function reaches a local maximum or minimum.
  2. Critical Points: Points where the derivative of a function is zero or undefined.
  3. First Derivative Test: A method to determine whether a critical point is a local maximum, minimum, or point of inflection.

#Summary and Key Takeaways

#Summary

  • The first derivative test is used to find and classify local extrema by analyzing the sign changes of the first derivative around critical points.
  • Critical points occur where the first derivative equals zero or is undefined.
  • The test involves checking the derivative values slightly to the left and right of the critical points.

#Key Takeaways

  • Local maxima occur where f′(x)f'(x)f′(x) changes from positive to negative.
  • Local minima occur where f′(x)f'(x)f′(x) changes from negative to positive.
  • Points of inflection occur where f′(x)f'(x)f′(x) does not change sign.

By mastering the first derivative test, you can efficiently identify and classify local extrema, a vital skill in calculus and many applications.

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Question 1 of 11

Ready to identify some critical points? 🤔 A critical point of a function f(x)f(x)f(x) is a point where:

f(x)=0f(x) = 0f(x)=0

f′(x)>0f'(x) > 0f′(x)>0

f′(x)=0f'(x) = 0f′(x)=0 or f′(x)f'(x)f′(x) is undefined

f′(x)<0f'(x) < 0f′(x)<0