Problem 21 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. $$ f(x)=5-2 x^{2}, \quad[0,3] $$

Step-by-Step Solution

Answer

The function \(f(x) = 5 - 2x^2\) has an absolute maximum value of 5 at \(x = 0\) and an absolute minimum value of -13 at \(x = 3\) within the given interval [0,3].

Step 1: Find the critical points

The critical points of a function are obtained by taking the derivative of the function and setting it equal to 0. Here, the derivative of \(f(x) = 5 - 2x^2\) is \(f'(x) = -4x\). Solving for \(f'(x) = 0\), we find that there is a critical point at \(x=0\).

Step 2: Compute the function values at critical point and endpoints

Compute the values of the function at the critical point and the endpoints. We get \(f(0) = 5\), \(f(3) = -13\) and \(f(0) = 5\) again as it's also the critical point.

Step 3: Identify the absolute extrema

Following from step 2, the absolute maximum value is 5 at \(x = 0\), while the absolute minimum value is -13 at \(x = 3\).

Step 4: Verify the results graphically

Plotting the function \(f(x)=5-2x^2\) on a graphing utility, one can confirm that the function indeed has a maximum point at \(x = 0\) and a minimum point at \(x = 3\) within the closed interval [0,3]. Thus, the calculated absolute extrema have been verified graphically.

Key Concepts

Critical Points

In the realm of calculus, critical points are fundamental in finding where a function reaches its highest or lowest points—known as extrema. Critical points occur where the derivative of a function is zero or undefined. For example, consider the function \(f(x) = 5 - 2x^2\). To find its critical points, we take the derivative \(f'(x) = -4x\). Set \(f'(x) = 0\) to find locations where the slope of the function is flat. Solving \(-4x = 0\), we get \(x = 0\). This indicates a potential extremum at \(x = 0\). Always remember, however, the critical point alone does not tell us if it is a maximum or minimum without further investigation.

Graphical Verification

Graphical verification is an essential step to confirming analytical results regarding extrema. Using graphing tools helps us visualize the function and verify calculated values. For the function \(f(x) = 5 - 2x^2\) over the interval [0,3], we graph it to see how it behaves. Visualizing the graph makes it easier to confirm our analytical findings. You will often see:

• The curve reaches its highest point at \(x = 0\), which matches our determined absolute maximum value.
• It reaches the lowest point at the endpoint \(x = 3\), consistent with our absolute minimum finding.

Graphical verification provides a visual confirmation, reinforcing confidence in our calculated results.

Endpoints of an Interval

When searching for absolute extrema on a closed interval, we cannot overlook the endpoints. Closed intervals, such as [0,3], include endpoint endpoints, which must be evaluated separately. Endpoint evaluations are crucial as they can potentially hold the global maximum or minimum values, especially when they are not critical points.For \(f(x) = 5 - 2x^2\) at the endpoints:

• At \(x = 0\), the function value is \(f(0) = 5\).
• At \(x = 3\), the function value is \(f(3) = -13\).

Compare these values with the critical point evaluations. Checking each helps ensure no possible extremum is missed. Thus, use endpoint evaluations, alongside critical point evaluations, to accurately determine global extrema within a closed interval.