Problem 38 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=5+54 x-2 x^{3}, \quad[0,4]$$

Step-by-Step Solution

Answer

The absolute maximum is 119 at \( x = 3 \), and the absolute minimum is 5 at \( x = 0 \).

Step 1: Find the derivative of the function

To find the extreme values of the function, we start by finding the derivative of the function. Given: \ \[ f(x) = 5 + 54x - 2x^3 \] \ The derivative of the function is: \ \[ f'(x) = \frac{d}{dx}(5 + 54x - 2x^3) = 54 - 6x^2 \]

Step 2: Set the derivative to zero to find critical points

To find the critical points of the function, set the derivative to zero and solve for \( x \): \ \[ 54 - 6x^2 = 0 \] \ Solving the equation gives: \ \[ 6x^2 = 54 \] \ \[ x^2 = 9 \] \ \[ x = 3 \] and \[ x = -3 \] (however, only \( x = 3 \) falls within the interval \([0, 4]\)).

Step 3: Evaluate the function at critical points and endpoints

Now evaluate \( f(x) \) at the critical point \( x = 3 \) and the endpoints of the interval \([0, 4]\). \ - For \( x = 0 \): \ \[ f(0) = 5 + 54(0) - 2(0)^3 = 5 \] \ - At the critical point \( x = 3 \): \ \[ f(3) = 5 + 54(3) - 2(3)^3 = 119 \] \ - For \( x = 4 \): \ \[ f(4) = 5 + 54(4) - 2(4)^3 = 101 \]

Step 4: Identify the absolute maximum and minimum values

Compare the function values to determine the absolute maximum and minimum: \ - \( f(0) = 5 \)- \( f(3) = 119 \) - \( f(4) = 101 \) \ The absolute maximum value is 119 at \( x = 3 \), and the absolute minimum value is 5 at \( x = 0 \).

Key Concepts

Derivative

To understand the process of finding extreme values, we first need to grasp the concept of a derivative. In simple terms, the derivative of a function gives us the rate at which the function's value changes with respect to changes in the input. It is like checking how fast or slow a certain value is changing. Derivatives are represented by the symbol \( f'(x) \) or \( \frac{d}{dx}f(x) \). For example, consider our function \( f(x) = 5 + 54x - 2x^3 \). By differentiating it, we find that \( f'(x) = 54 - 6x^2 \). This tells us how the output of the function is changing as \( x \) changes. It's crucial in identifying where the function might reach a peak or valley, which leads us to critical points.

Critical Points

Critical points are the x-values where the derivative of a function equals zero or is undefined. These points may indicate potential maxima or minima—places where the function could change directions (from increasing to decreasing or vice versa). In our example, once we've calculated \( f'(x) = 54 - 6x^2 \), we set it to zero to find the critical points:

• \( 54 - 6x^2 = 0 \)
• Solve for \( x \): \( x^2 = 9 \)
• \( x = 3 \) and \( x = -3 \)

However, since our interval is \([0, 4]\), \( x = -3 \) falls outside of this range, leaving us with the critical point \( x = 3 \). By examining these points further, we can determine which of these might lead to the function's highest or lowest values over a given range.

Absolute Maximum and Minimum

By evaluating a function at its critical points and the endpoints of the given interval, we can determine its absolute maximum and minimum. This means finding the highest and lowest values of the function within a specified range. From our earlier steps, we found the critical point \( x = 3 \) within the interval \([0, 4]\). Then, we evaluated the function at:

• Start of the interval, \( x = 0 \): \( f(0) = 5 \)
• Critical point, \( x = 3 \): \( f(3) = 119 \)
• End of the interval, \( x = 4 \): \( f(4) = 101 \)

After comparing these values, we see that the absolute maximum value of 119 occurs at \( x = 3 \), and the absolute minimum value of 5 occurs at \( x = 0 \). These points provide the function's peak and lowest point within our given interval.

Interval Evaluation

Interval evaluation is essential when determining a function’s extreme values. It involves not only solving for and evaluating the function at critical points but also checking the endpoints of the interval.The extreme value theorem suggests that if a function is continuous on a closed interval, like \([0, 4]\), it will have both a maximum and minimum value. To find these values, we:

• Look for critical points where the derivative equals zero or is undefined
• Calculate the function's value at these and at the interval's endpoints

By performing these evaluations, we ensure no potential maximum or minimum values are missed. In our case, by evaluating \( f(x) \) at \( x = 0 \), \( x = 3 \), and \( x = 4 \), we confirm the highest and lowest function values within the given limits.