Math Problem Statement

Given that g(x) is continuous for all real values and the graph of g'(x) is given below, find all the critical values of g(x) and use the first derivative test to determine whether g(x) has a local minimum, local maximum, or neither at each critical value.

Solution

This problem asks us to analyze the graph of g(x)g'(x), the first derivative of g(x)g(x), to find the critical points and determine the nature of g(x)g(x) (local maximum, minimum, or neither) at each of these critical values.

Step-by-step approach:

  1. Identify Critical Points:

    • Critical points occur where g(x)=0g'(x) = 0 or where g(x)g'(x) is undefined. In this case, look for:
      • Where the graph of g(x)g'(x) crosses the x-axis (i.e., g(x)=0g'(x) = 0).
      • Any discontinuities in the graph of g(x)g'(x).
  2. Examine the Graph:

    • The graph of g(x)g'(x) shows three key points where g(x)=0g'(x) = 0 (where the graph intersects the x-axis):
      • At x=2x = -2
      • At x=1x = 1
      • At x=4x = 4
    • There is also a discontinuity at x=0x = 0, where the graph "jumps."
  3. Use the First Derivative Test:

    • To determine the nature of each critical point (local maximum, minimum, or neither), check how g(x)g'(x) changes sign around each critical point:
      • At x=2x = -2: g(x)g'(x) changes from positive (to the left of x=2x = -2) to negative (to the right of x=2x = -2). This indicates a local maximum at x=2x = -2.
      • At x=1x = 1: g(x)g'(x) changes from negative to positive, indicating a local minimum at x=1x = 1.
      • At x=4x = 4: g(x)g'(x) does not change sign (it remains positive), so this is neither a maximum nor a minimum.
      • At x=0x = 0: There's a discontinuity. Since we cannot determine the behavior of g(x)g'(x) at this point using the first derivative test, this point requires special attention. Given the nature of g(x)g'(x), g(x)g(x) may not have an extremum at x=0x = 0.

Conclusion:

  • Local maximum at x=2x = -2.
  • Local minimum at x=1x = 1.
  • Neither at x=4x = 4 and x=0x = 0.

Would you like further explanation or details on any of these steps?

Related questions:

  1. What is the significance of the first derivative in identifying local extrema?
  2. How does the second derivative test differ from the first derivative test?
  3. Why is g(x)g(x) continuous even though g(x)g'(x) is not at certain points?
  4. Can we always use the first derivative test for non-continuous derivative graphs?
  5. What other methods can be used to confirm extrema besides the first derivative test?

Tip:

When analyzing the first derivative graph, always check for sign changes around critical points, as they provide key information about the behavior of the original function.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
First Derivative Test

Formulas

g'(x) = 0 for critical points
Sign change of g'(x) determines max/min

Theorems

First Derivative Test
Continuity of g(x)

Suitable Grade Level

Grades 11-12, Calculus I