Problem 97 - University Calculus: Early Transcendentals

Question

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=\pi x^{2} e^{-3 x / 2}, \quad[0,5]$$

Step-by-Step Solution

Answer

The absolute maximum is at \(x = \frac{4}{3}\) and the minimum is at \(x = 0\).

Step 1: Plot the Function

We begin by plotting the function \( f(x) = \pi x^{2} e^{-3x/2} \) over the interval \([0,5]\). This helps us visualize how the function behaves within this range. A computer algebra system (CAS) can be used to generate this plot, which should show the decrease of the function as \(x\) increases.

Step 2: Find Points Where f' = 0

First, we need to find the derivative of the function: \( f'(x) = \pi \left(2x e^{-3x/2} - \frac{3x^2}{2} e^{-3x/2}\right) \). We set \( f'(x) = 0 \) and solve for \( x \). Simplifying, it gives: \( 2x - \frac{3x^2}{2} = 0 \). Solving this equation numerically (or analytically using the quadratic formula if appropriate) will give us critical points within \( [0,5] \).

Step 3: Solve f'(x) = 0

Continuing from Step 2, we solve \( 2x - \frac{3x^2}{2} = 0 \). Rewriting it gives \( 4x = 3x^2 \) or \( x = 0 \) and \( x = \frac{4}{3} \). These are the critical points where \( f'(x) = 0 \) within our interval.

Step 4: Find Points Where f' Does Not Exist

To find where \( f'(x) \) does not exist, we inspect the expression of \( f'(x) \). There are no terms that lead to division by zero or other undefined operations within the interval \([0,5]\). Therefore, \( f'(x) \) exists everywhere in this interval.

Step 5: Evaluate f(x) at Key Points

Now we evaluate \( f(x) = \pi x^{2} e^{-3x/2} \) at the critical points \( x = 0 \), \( x = \frac{4}{3} \), and the endpoints \( x = 0 \) and \( x = 5 \). - \( f(0) = \pi \times 0^2 \times e^{0} = 0 \)- \( f(\frac{4}{3}) = \pi (\frac{4}{3})^2 e^{-2} \approx 1.160705 \)- \( f(5) = \pi \times 5^2 \times e^{-15/2} \approx 0.221215 \)

Step 6: Identify Absolute Extrema

Comparing the above values, the maximum value is approximately \(1.160705\) at \(x = \frac{4}{3}\) and the minimum value is \(0\) at \(x = 0\). Therefore, the absolute maximum occurs at \(x = \frac{4}{3}\) and the minimum occurs at \(x = 0\).

Key Concepts

Critical Points

Critical points are significant values in a mathematical function where its derivative equals zero. Finding these points allows us to understand where a function could potentially reach a maximum or minimum value. To identify critical points, we take the derivative of the function and set it equal to zero. This is where the slope of the tangent line to the function is flat. For instance, if we consider the function

• \( f(x) = \pi x^{2} e^{-3x/2} \),

To find its critical points, we first compute its derivative, denoted as \( f'(x) \). If we set \( f'(x) = 0 \), we would solve for the variable \( x \) to locate these points within the given interval. Identifying critical points helps in assessing parts of the function where important changes like peaks or troughs occur. This is crucial for subsequent steps like evaluating absolute extrema.

Derivative

The derivative of a function reflects its rate of change. It essentially tells us how the function's value changes as we adjust an input value (i.e., how steep or flat the curve is at any point). The derivative is a foundational concept in calculus and is denoted as \( f'(x) \) when contemplating a function \( f(x) \).

• In the problem given, the derivative \( f'(x) \) of \( f(x) = \pi x^{2} e^{-3x/2} \) was calculated as \ \( \pi \left(2x e^{-3x/2} - \frac{3x^2}{2} e^{-3x/2}\right) \).
• Derivatives are crucial for discovering both the steepness of a curve and for finding critical points – as we set the derivative to zero and solve to find where these points occur.
• Additionally, investigating where a derivative does not exist is key.

The absence of a derivative at a certain point could signify a cusp or a sharp change in the function's graph, which might also represent local extremes in a function.

Absolute Extrema

Absolute extrema are the highest and lowest points a function achieves over a specified interval. For our interval

• In the exercise, we examined both the critical points and the function's endpoints to determine absolute extrema.
• For example, evaluating \( f(x) \) at its critical points and the boundaries [0,5] helped identify
  • the absolute maximum: approximately \(1.160705\) at \(x = \frac{4}{3}\),
  • and the absolute minimum: 0 at \(x = 0\).

The computed values provide insight into the function’s overall behavior and give insight into real-world applications where extreme values might represent optimal conditions or limits.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a software tool used for solving mathematical problems and performing symbolical computations such as deriving, integrating, and plotting functions. These systems are incredibly beneficial for students and researchers who need to handle complex mathematical calculations efficiently.

• In this exercise, a CAS was employed to assist in plotting the function \( f(x) \) and its derivative within the specified interval.
• Using a CAS, users input a function and receive outputs such as plots, numerical solutions, and symbolic derivatives. These are invaluable for cross-verification and reducing manual calculation errors.

Employing a CAS reduces the time and effort spent on mathematical problem-solving, helping learners focus more on understanding and applying concepts effectively.