Find The Largest And Smallest Values ​​of The Function F(x).

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In calculus, a crucial task involves determining the maximum and minimum values of a given function within a specified interval. This process, known as optimization, has wide-ranging applications in various fields, from engineering and economics to computer science and data analysis. In this comprehensive guide, we will delve into the methods for finding extreme values of functions, illustrating the concepts with practical examples and step-by-step solutions.

Understanding Extreme Values

Before we embark on the journey of finding maximum and minimum values, let's first grasp the fundamental concepts. A function's extreme values refer to its highest and lowest points within a given interval. These extreme values can be either absolute (global) or local (relative). An absolute maximum is the highest value the function attains over the entire interval, while an absolute minimum is the lowest value. On the other hand, a local maximum is a point where the function's value is greater than its neighbors, and a local minimum is a point where the function's value is lower than its neighbors.

Finding extreme values is essential for various applications. For example, engineers may need to determine the maximum stress a bridge can withstand, while economists might want to find the production level that maximizes profit. Understanding the concept of extreme values empowers us to solve real-world problems effectively.

Methods for Finding Extreme Values

Calculus provides us with powerful tools to identify extreme values of functions. The two primary methods we will explore are:

  1. The First Derivative Test: This method involves analyzing the sign of the function's first derivative to identify critical points, which are potential locations of extreme values.
  2. The Second Derivative Test: This method utilizes the second derivative to determine the nature of critical points, distinguishing between local maxima, local minima, and saddle points.

Let's delve into each method in detail, accompanied by illustrative examples.

The First Derivative Test

The first derivative of a function, denoted as f'(x), provides valuable information about the function's rate of change. At a critical point, where the derivative is either zero or undefined, the function's slope is either horizontal or vertical, indicating a potential extreme value. The first derivative test leverages this principle to locate and classify critical points.

Steps Involved in the First Derivative Test:

  1. Find the derivative: Compute the first derivative of the function, f'(x).
  2. Find critical points: Set f'(x) = 0 and solve for x to find the critical points. Also, identify any points where f'(x) is undefined.
  3. Create a sign chart: Divide the domain of the function into intervals based on the critical points. Choose a test value within each interval and evaluate f'(x) at that value. Record the sign of f'(x) in a sign chart.
  4. Analyze the sign chart:
    • If f'(x) changes from positive to negative at a critical point, then the function has a local maximum at that point.
    • If f'(x) changes from negative to positive at a critical point, then the function has a local minimum at that point.
    • If f'(x) does not change sign at a critical point, then the function has neither a local maximum nor a local minimum at that point.
  5. Evaluate the function: Evaluate the function f(x) at each critical point and at the endpoints of the interval to determine the absolute maximum and minimum values.

To illustrate the first derivative test, let's consider the function:

  • f(x) = x^3 - 3x^2 + 2

We aim to find the extreme values of this function on the interval [-1, 3].

  1. Find the derivative: f'(x) = 3x^2 - 6x
  2. Find critical points:
    • Set f'(x) = 0: 3x^2 - 6x = 0
    • Factor: 3x(x - 2) = 0
    • Solve for x: x = 0 or x = 2
    • The critical points are x = 0 and x = 2.
  3. Create a sign chart:
Interval Test Value f'(x) Sign Behavior
(-1, 0) -0.5 3.75 + Increasing
(0, 2) 1 -3 - Decreasing
(2, 3) 2.5 3.75 + Increasing
  1. Analyze the sign chart:
    • At x = 0, f'(x) changes from positive to negative, indicating a local maximum.
    • At x = 2, f'(x) changes from negative to positive, indicating a local minimum.
  2. Evaluate the function:
    • f(-1) = (-1)^3 - 3(-1)^2 + 2 = -2
    • f(0) = 0^3 - 3(0)^2 + 2 = 2
    • f(2) = 2^3 - 3(2)^2 + 2 = -2
    • f(3) = 3^3 - 3(3)^2 + 2 = 2

Therefore, the absolute maximum value of the function on the interval [-1, 3] is 2, which occurs at x = 0 and x = 3. The absolute minimum value is -2, which occurs at x = -1 and x = 2.

The Second Derivative Test

The second derivative of a function, denoted as f''(x), provides information about the concavity of the function. A positive second derivative indicates that the function is concave up, while a negative second derivative indicates that the function is concave down. The second derivative test utilizes this principle to classify critical points.

Steps Involved in the Second Derivative Test:

  1. Find the first derivative: Compute the first derivative of the function, f'(x).
  2. Find critical points: Set f'(x) = 0 and solve for x to find the critical points. Also, identify any points where f'(x) is undefined.
  3. Find the second derivative: Compute the second derivative of the function, f''(x).
  4. Evaluate the second derivative at critical points: Evaluate f''(x) at each critical point.
  5. Analyze the second derivative:
    • If f''(x) > 0 at a critical point, then the function has a local minimum at that point.
    • If f''(x) < 0 at a critical point, then the function has a local maximum at that point.
    • If f''(x) = 0 at a critical point, then the test is inconclusive, and we need to use the first derivative test.
  6. Evaluate the function: Evaluate the function f(x) at each critical point and at the endpoints of the interval to determine the absolute maximum and minimum values.

Let's illustrate the second derivative test with the same function we used for the first derivative test:

  • f(x) = x^3 - 3x^2 + 2
  1. Find the first derivative: f'(x) = 3x^2 - 6x
  2. Find critical points:
    • Set f'(x) = 0: 3x^2 - 6x = 0
    • Factor: 3x(x - 2) = 0
    • Solve for x: x = 0 or x = 2
    • The critical points are x = 0 and x = 2.
  3. Find the second derivative: f''(x) = 6x - 6
  4. Evaluate the second derivative at critical points:
    • f''(0) = 6(0) - 6 = -6
    • f''(2) = 6(2) - 6 = 6
  5. Analyze the second derivative:
    • At x = 0, f''(0) < 0, indicating a local maximum.
    • At x = 2, f''(2) > 0, indicating a local minimum.
  6. Evaluate the function:
    • f(-1) = (-1)^3 - 3(-1)^2 + 2 = -2
    • f(0) = 0^3 - 3(0)^2 + 2 = 2
    • f(2) = 2^3 - 3(2)^2 + 2 = -2
    • f(3) = 3^3 - 3(3)^2 + 2 = 2

The second derivative test confirms the results obtained from the first derivative test: the absolute maximum value is 2, occurring at x = 0 and x = 3, and the absolute minimum value is -2, occurring at x = -1 and x = 2.

H2: Applying Extreme Value Techniques to Specific Functions

Now, let's put our knowledge into practice by tackling the specific functions presented in the original prompt. We will demonstrate how to find the maximum and minimum values of these functions on the given intervals.

H3: Function a: f(x) = x^4 - 8x^2 - 9

We are tasked with finding the extreme values of the function f(x) = x^4 - 8x^2 - 9 on the intervals [-1, 1] and [0, 3]. Let's systematically apply our methods.

Interval [-1, 1]:

  1. Find the first derivative:
    • f'(x) = 4x^3 - 16x
  2. Find critical points:
    • Set f'(x) = 0: 4x^3 - 16x = 0
    • Factor: 4x(x^2 - 4) = 0
    • Solve for x: x = 0, x = -2, x = 2
    • The critical points are x = 0, x = -2, and x = 2. However, only x = 0 and the interval's endpoints, x = -1 and x = 1, fall within the interval [-1, 1].
  3. Find the second derivative:
    • f''(x) = 12x^2 - 16
  4. Evaluate the second derivative at critical points:
    • f''(0) = 12(0)^2 - 16 = -16 (Local Maximum)
  5. Evaluate the function at critical points and endpoints:
    • f(-1) = (-1)^4 - 8(-1)^2 - 9 = -16
    • f(0) = (0)^4 - 8(0)^2 - 9 = -9
    • f(1) = (1)^4 - 8(1)^2 - 9 = -16

Therefore, on the interval [-1, 1], the absolute maximum value of f(x) is -9 at x = 0, and the absolute minimum value is -16 at x = -1 and x = 1.

Interval [0, 3]:

  1. Critical Points within [0, 3]: From our previous calculation, the critical points are x = 0, x = -2, and x = 2. Within the interval [0, 3], we consider x = 0 and x = 2.
  2. Evaluate the function at critical points and endpoints:
    • f(0) = (0)^4 - 8(0)^2 - 9 = -9
    • f(2) = (2)^4 - 8(2)^2 - 9 = 16 - 32 - 9 = -25
    • f(3) = (3)^4 - 8(3)^2 - 9 = 81 - 72 - 9 = 0

Thus, on the interval [0, 3], the absolute maximum value of f(x) is 0 at x = 3, and the absolute minimum value is -25 at x = 2.

H3: Function b: f(x) = (x^2 + 4) / x

Next, we'll determine the extreme values for the function f(x) = (x^2 + 4) / x on the intervals [-4, -1] and [1, 3].

Interval [-4, -1]:

  1. Find the first derivative:
    • f'(x) = (x(2x) - (x^2 + 4)(1)) / x^2 = (2x^2 - x^2 - 4) / x^2 = (x^2 - 4) / x^2
  2. Find critical points:
    • Set f'(x) = 0: (x^2 - 4) / x^2 = 0
    • Solve for x: x^2 - 4 = 0 => x = -2, x = 2
    • x = 0 is also a critical point where f'(x) is undefined, but it's not in the domain of f(x).
    • Within the interval [-4, -1], only x = -2 is relevant.
  3. Find the second derivative:
    • f''(x) = ((x^2)(2x) - (x^2 - 4)(2x)) / x^4 = (2x^3 - 2x^3 + 8x) / x^4 = 8 / x^3
  4. Evaluate the second derivative at critical points:
    • f''(-2) = 8 / (-2)^3 = -1 (Local Maximum)
  5. Evaluate the function at critical points and endpoints:
    • f(-4) = ((-4)^2 + 4) / -4 = -5
    • f(-2) = ((-2)^2 + 4) / -2 = -4
    • f(-1) = ((-1)^2 + 4) / -1 = -5

Therefore, on the interval [-4, -1], the absolute maximum value of f(x) is -4 at x = -2, and the absolute minimum value is -5 at x = -4 and x = -1.

Interval [1, 3]:

  1. Critical Points within [1, 3]: From our previous calculation, the critical points are x = -2, x = 2. Within the interval [1, 3], only x = 2 is relevant.
  2. Evaluate the function at critical points and endpoints:
    • f(1) = ((1)^2 + 4) / 1 = 5
    • f(2) = ((2)^2 + 4) / 2 = 4
    • f(3) = ((3)^2 + 4) / 3 = 13 / 3 ≈ 4.33

Thus, on the interval [1, 3], the absolute maximum value of f(x) is 5 at x = 1, and the absolute minimum value is 4 at x = 2.

H2: Conclusion

In this comprehensive guide, we have explored the methods for finding the maximum and minimum values of functions, emphasizing the first and second derivative tests. By applying these techniques systematically, we can effectively identify extreme values and solve optimization problems in various domains. The step-by-step solutions provided for the specific functions serve as a practical illustration of the concepts discussed, empowering readers to tackle similar problems with confidence.

By mastering the art of finding extreme values, you unlock a powerful tool for optimizing processes, making informed decisions, and gaining deeper insights into the behavior of functions.