Find The Exact Location Of All The Relative And Absolute Extrema Of The Function. (order Your Answers

Find The Exact Location Of All The Relative And Absolute Extrema Of The Function. (order Your Answers

Understanding how to identify the relative and absolute extrema of a function is a fundamental aspect of calculus and mathematical analysis. These extrema—comprising local maxima and minima as well as the global (absolute) maximum and minimum—are critical points that reveal the behavior of the function, informing decisions in engineering, economics, physics, and various applied sciences. This comprehensive guide will walk you through the systematic process of locating these extrema, emphasizing proper order, accuracy, and clarity in your answers.

Introduction to Extrema of Functions

Extrema refer to points on a function where the function attains a maximum or minimum value within a certain interval or over its entire domain. These points are characterized by specific properties related to derivatives, which enable us to locate them analytically.

Types of Extrema

• Relative (Local) Extrema: Points where the function reaches a local maximum or minimum in a neighborhood around the point.
• Absolute (Global) Extrema: Points where the function attains its highest or lowest value over the entire domain.

Step-by-Step Procedure to Find Extrema

To accurately find all the relative and absolute extrema, follow this structured process:

1. Determine the Domain and Critical Points

• Identify the domain: Clearly define the interval or set over which the function is analyzed.
1. Find critical points: Calculate where the derivative of the function equals zero or does not exist within the domain.
   • Set the first derivative \(f'(x)\) equal to zero: \(f'(x) = 0\)
   • Find points where \(f'(x)\) is undefined, provided these points are within the domain.

2. Analyze Critical Points Using the First Derivative Test

• Determine the sign of \(f'(x)\) immediately to the left and right of each critical point.
1. Classify each critical point:
   • If \(f'\) changes from positive to negative, the critical point is a local maximum.
   • If \(f'\) changes from negative to positive, the critical point is a local minimum.
   • If \(f'\) does not change sign, the point is neither a max nor a min (possible inflection point).

3. Use the Second Derivative Test (if applicable)

• Calculate the second derivative \(f''(x)\) at each critical point.
1. Classify the critical points:
   • If \(f''(x) > 0\), the point is a local minimum.
   • If \(f''(x) < 0\), the point is a local maximum.
   • If \(f''(x) = 0\), the test is inconclusive; consider higher-order derivatives or alternative methods.

4. Find Absolute Extrema

• Evaluate the function at all critical points identified.
• Evaluate the function at the endpoints of the domain if it is a closed interval.
1. Compare all these values:
   • The largest value among these is the absolute maximum.
   • The smallest value among these is the absolute minimum.

Detailed Example with Explanation

Let's illustrate the process with a comprehensive example to demonstrate each step effectively.

Example Function: \(f(x) = x^3 - 6x^2 + 9x + 2\)

Step 1: Determine the Domain and Critical Points

• Assume the domain is all real numbers (\(-\infty, \infty\)).


• Calculate the first derivative:

\[
f'(x) = 3x^2 - 12x + 9
\]

• Find critical points by solving \(f'(x) = 0\):

\[
3x^2 - 12x + 9 = 0
\]
\[
x^2 - 4x + 3 = 0
\]
\[
(x - 1)(x - 3) = 0
\]

• Critical points at \(x = 1\) and \(x = 3\).

Step 2: Analyze Critical Points Using the First Derivative Test

• Choose test points around \(x=1\) and \(x=3\):


• For \(x < 1\), say \(x=0\):

\[
f'(0) = 0 - 0 + 9 = 9 > 0
\]

• For \(1 < x < 3\), say \(x=2\):

\[
f'(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 < 0
\]

• For \(x > 3\), say \(x=4\):

\[
f'(4) = 3(16) - 12(4) + 9 = 48 - 48 + 9 = 9 > 0
\]

• Sign Changes:


• At \(x=1\): \(f'\) changes from positive to negative → local maximum.


• At \(x=3\): \(f'\) changes from negative to positive → local minimum.

Step 3: Use the Second Derivative Test

• Calculate the second derivative:

\[
f''(x) = 6x - 12
\]

• At \(x=1\):

\[
f''(1) = 6(1) - 12 = -6 < 0 \Rightarrow\) local maximum at \(x=1\).

• At \(x=3\):

\[
f''(3) = 6(3) - 12 = 6 > 0 \Rightarrow\) local minimum at \(x=3\).

Step 4: Find the Exact Extrema Values

• Evaluate \(f(x)\) at critical points:


• \(f(1) = (1)^3 - 6(1)^2 + 9(1) + 2 = 1 - 6 + 9 + 2 = 6\)


• \(f(3) = (3)^3 - 6(3)^2 + 9(3) + 2 = 27 - 54 + 27 + 2 = 2\)


• Since the domain is all real numbers, check behavior as \(x \to \pm \infty\):


• As \(x \to \infty\), \(f(x) \to \infty\).


• As \(x \to -\infty\), \(f(x) \to -\infty\).


• Therefore:


• Absolute maximum: the function tends to infinity as \(x \to \infty\), so no finite absolute maximum exists.


• Absolute minimum: since the function tends to \(-\infty\) as \(x \to -\infty\), no finite absolute minimum exists.

Summary of Extrema:

• Local maximum at \((1, 6)\).


• Local minimum at \((3, 2)\).


• No finite absolute extrema over \(\mathbb{R}\).

Special Considerations for Different Domains

The process above assumes an unbounded domain. However, in many problems, the domain is restricted, such as a closed interval \([a, b]\). In such cases:

• Always evaluate the function at the endpoints \(a\) and \(b\).


• Critical points within the interval are analyzed as before.


• The absolute extrema are among the critical points and endpoints.

Example: Finding Extrema on a Closed Interval

Suppose \(f(x) = x^3 - 6x^2 + 9x + 2\) over \([0, 4]\).

• Critical points within \([0, 4]\): \(x=1\) and \(x=3\).


• Evaluate at critical points:


• \(f(1) = 6\)


• \(f(3) = 2\)


• Evaluate at endpoints:


• \(f(0) = 0 - 0 + 0 + 2 = 2\)


• \(f(4) = 64 - 96 + 36 + 2 = 6\)


• Determine extrema:


• Absolute maximum: