Problem 8 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. \(f(x)=8 x-x^{2} ;[0,6]\)

Step-by-Step Solution

Answer

Absolute maximum is 16 at \(x = 4\), minimum is 0 at \(x = 0\).

Step 1: Find the Critical Points

To find the critical points, we need the derivative of the function. The derivative of \(f(x) = 8x - x^2\) is \(f'(x) = 8 - 2x\). We set the derivative equal to zero to find the critical points: \(8 - 2x = 0\). Solving for \(x\), we get \(x = 4\). So, the critical point is \(x = 4\).

Step 2: Evaluate the Function at Critical Points and Endpoints

We evaluate the function \(f(x) = 8x - x^2\) at the critical point and the endpoints of the interval \([0, 6]\). Let's calculate:- At \(x = 0\): \(f(0) = 8(0) - 0^2 = 0\).- At \(x = 4\): \(f(4) = 8(4) - 4^2 = 32 - 16 = 16\).- At \(x = 6\): \(f(6) = 8(6) - 6^2 = 48 - 36 = 12\).

Step 3: Identify the Absolute Maximum and Minimum Values

The function values we obtained are \(f(0) = 0\), \(f(4) = 16\), and \(f(6) = 12\). The absolute maximum value is 16 at \(x = 4\), and the absolute minimum value is 0 at \(x = 0\).

Key Concepts

Critical Points

Understanding critical points is essential when determining the maximum and minimum of a function on a closed interval. Critical points are the points on a graph where the derivative is either zero or undefined. These points indicate where the slope of the tangent is horizontal or where a corner or a cusp occurs.

To find a critical point, follow these steps:

• Compute the derivative of the function. This helps identify where the slope changes.
• Set the derivative equal to zero and solve for the variable. This will give you potential points where the function has maxima (peaks) or minima (valleys).

In our example, the function is given by: \[ f(x) = 8x - x^2 \]The derivative is: \[ f'(x) = 8 - 2x \]Setting this equal to zero, \( 8 - 2x = 0 \), we solve \( x = 4 \). This is the critical point within the interval, indicating a potential location for either a maximum or minimum value of the function.

Derivative

A derivative represents how a function changes as its input changes. It is a fundamental concept in calculus used to investigate the rate of change of quantities. In simpler terms, it helps us understand how steep a curve is at any given point.

Here’s how you determine the derivative of a function:

• Use differentiation rules, such as the power rule, product rule, or chain rule, to find the slope of the tangent line to a curve at any given point.

For the function \( f(x) = 8x - x^2 \), apply the derivative rules:

• The derivative of \( 8x \) is 8.
• The derivative of \( -x^2 \) is \( -2x \) (applying the power rule).

Thus, the derivative of the function is:\[ f'(x) = 8 - 2x \]By interpreting this derivative, we can then locate the critical points, where maximal or minimal changes occur in the original function.

Function Evaluation

Function evaluation is the process of finding the output of a function given an input value. It’s a necessary step to determine the actual maximum and minimum values of a function within a given interval.

To evaluate the function at any given point:

• Simply plug the value into the function's formula.
• Perform the arithmetic operations as indicated.

In our example, the critical point is at \( x = 4 \), and we are also tasked with evaluating the function at the endpoints of the interval, \( x = 0 \) and \( x = 6 \).

Calculating the function values:

• At \( x = 0 \): \( f(0) = 8(0) - 0^2 = 0 \)
• At \( x = 4 \): \( f(4) = 8(4) - 4^2 = 32 - 16 = 16 \)
• At \( x = 6 \): \( f(6) = 8(6) - 6^2 = 48 - 36 = 12 \)

These evaluations indicate that the absolute maximum is 16 at \( x = 4 \) and the absolute minimum is 0 at \( x = 0 \). This demonstrates the importance of evaluating the function at critical points and interval boundaries when finding extrema.