Problem 37 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=12+4 x-x^{2}, \quad[0,5]$$

Step-by-Step Solution

Answer

The absolute maximum is 16 at \(x = 2\), and the absolute minimum is 7 at \(x = 5\).

Step 1: Identify the critical points

To find the critical points, we first find the derivative of the function. The derivative of \( f(x) = 12 + 4x - x^2 \) is \( f'(x) = 4 - 2x \). Set the derivative equal to zero to find the critical points: \[ 4 - 2x = 0 \]Solve for \(x\): \[ x = 2 \].

Step 2: Evaluate the function at the critical point

Next, evaluate the function \( f(x) = 12 + 4x - x^2 \) at the critical point we found: \( f(2) = 12 + 4(2) - (2)^2 = 12 + 8 - 4 = 16 \).

Step 3: Evaluate the function at the endpoints

Evaluate the function at the endpoints of the interval \([0, 5]\):- At \( x = 0 \): \( f(0) = 12 + 4(0) - (0)^2 = 12 \).- At \( x = 5 \): \( f(5) = 12 + 4(5) - (5)^2 = 12 + 20 - 25 = 7 \).

Step 4: Determine the absolute maximum and minimum

Compare the values from Steps 2 and 3. The values obtained were:- \( f(0) = 12 \)- \( f(2) = 16 \)- \( f(5) = 7 \)The absolute maximum value is 16 at \( x = 2 \), and the absolute minimum value is 7 at \( x = 5 \).

Key Concepts

Critical Points

Critical points are important when we look for extreme values of a function, such as maximum or minimum values over an interval. If a function's derivative is zero or undefined at a certain point, that point is a critical point. This is because the slope of the tangent at that point is horizontal, hinting that there could be a peak or a valley in the graph of the function.
To find critical points:

• Begin by finding the derivative of the function.
• Set the derivative equal to zero and solve for the variable.

For example, for the function \( f(x) = 12 + 4x - x^2 \), its derivative \( f'(x) = 4 - 2x \) shows horizontal slopes where \( 4 - 2x = 0 \), leading us to the critical point \( x = 2 \). This method allows us to locate where the function's behavior might change.

Derivative

The derivative is crucial in calculus for understanding the behavior of functions. It represents the rate of change or the slope of a function. In simpler terms, it tells us how steep the function is at any given point. Knowing this helps us identify where the function increases or decreases.
For the function \( f(x) = 12 + 4x - x^2 \), we calculate the derivative as follows:

• The derivative of a constant like 12 is 0.
• The derivative of \( 4x \) is simply 4, as a linear term's derivative leaves only the coefficient.
• The derivative of \( -x^2 \) is \(-2x\).

Combining these, the derivative \( f'(x) = 4 - 2x \) clearly demonstrates how the function changes. With this derivative, we find critical points by setting \( f'(x) \) to zero.

Function Evaluation

Function evaluation is the process of calculating the value of a function at specific points. It helps determine how a function behaves at critical points and other significant points like interval endpoints. Evaluating a function tells you the actual output of the function for a particular input.
In our exercise, to find the absolute extrema, we evaluated the function at the critical point \( x = 2 \) and at the interval endpoints \([0, 5]\):

• At \( x = 2 \), the function value is \( f(2) = 16 \).
• At \( x = 0 \), the function value is \( f(0) = 12 \).
• At \( x = 5 \), the function value is \( f(5) = 7 \).

These values helped us determine which points represented the absolute maximum and minimum of the function within the given interval.

Interval Endpoints

When searching for absolute extrema, evaluating the function at the interval's endpoints is as important as evaluating it at the critical points. Endpoints can often hold the absolute minimum or maximum value of a function over a closed interval.
The given function is evaluated over the interval \([0, 5]\) meaning the endpoints are \( x = 0 \) and \( x = 5 \). The function values at these points were:

• \( f(0) = 12 \)
• \( f(5) = 7 \)

Comparing all the values obtained, the endpoint \( x = 5 \) provided the absolute minimum value of 7, whereas a non-endpoint, \( x = 2 \), held the absolute maximum value of 16. Without evaluating the endpoints, you might miss important extrema!