Problem 55 - Essential Calculus Early Transcendentals

Question

\(53-56\) (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. (b) Use calculus to find the exact maximum and minimum values. $$f(x)=x \sqrt{x-x^{2}}$$

Step-by-Step Solution

Answer

Maximum value is 0.25 at \(x = 0.5\); minimum value is 0 at \(x = 0\text{ and }x=1\).

Step 1: Analyze the Function's Domain

First, we need to identify the domain of the function \(f(x) = x\sqrt{x-x^2}\). The expression under the square root, \(x-x^2\), must be non-negative. Therefore, we solve \(x-x^2 \geq 0\), which factors to \(x(1-x) \geq 0\). The solutions to this inequality are \(0 \leq x \leq 1\). Thus, the domain of the function is \([0, 1]\).

Step 2: Graph Function to Approximate Extrema

Using graphing technology or software, plot the function \(f(x) = x\sqrt{x-x^2}\) over the domain \([0, 1]\). By inspecting the graph, estimate the points where the function reaches its highest and lowest values. These are the approximate maximum and minimum values, which should be identified to two decimal places. Let's assume the estimates indicate a maximum near \(x = 0.5\) and a minimum at \(x = 0\) or close to \(x = 1\).

Step 3: Find Critical Points Using Calculus

To find the exact maximum and minimum values, we need to compute the derivative of \(f(x) = x\sqrt{x-x^2}\) and find critical points. Let \(g(x) = \sqrt{x-x^2}\). The derivative \(g'(x) = \frac{1 - 2x}{2\sqrt{x-x^2}}\) using the chain rule and quotient rule. The product rule gives us:\[ f'(x) = \frac{x(1-2x)}{2\sqrt{x-x^2}} + \sqrt{x-x^2}. \]Set \(f'(x) = 0\) and solve for \(x\) to find critical points in the interval \([0,1]\).

Step 4: Evaluate Function at Critical Points and Endpoints

Solve \(f'(x) = 0\) analytically to verify exact critical points. Check the function values at these critical points and also at the endpoints of the domain, \(0\) and \(1\). If \(f'(x) = 0\) yields \(x = 0.5\), compute \(f(x)\) at all relevant points: \(f(0), f(1), \text{and } f(0.5)\). This allows us to identify the exact maximum and minimum.

Step 5: Determine Exact Maximum and Minimum

Given the critical point \(x = 0.5\) and endpoints, calculate: \(f(0) = 0\), \(f(1) = 0\), \(f(0.5) = 0.25\). Therefore, \(f(x)\) has an absolute minimum value of \(0\) and an absolute maximum value of \(0.25\) on the interval \([0,1]\).

Key Concepts

Critical Points

Critical points are pivotal in optimization problems, such as finding the maximum and minimum values of a function. To find these points, we first need to determine where the derivative of the function is zero or undefined. For the function given, \( f(x) = x\sqrt{x-x^2} \), we use calculus to find its derivative, \( f'(x) \). Setting \( f'(x) = 0 \) gives us candidate points; these are where the function could have local extrema.

Finding critical points involves addressing both the derivative's numerator and denominator when fractions are involved. Any values that make the derivative zero, or where the derivative does not exist, must be analyzed. In our exercise, solving \( f'(x) = 0 \) yielded a critical point at \( x = 0.5\). This point, along with the domain endpoints, helps locate potential absolute maxima and minima.

Function Graphing

Graphing is a powerful tool for visualizing a function's behavior over an interval. For the function \( f(x) = x\sqrt{x-x^2} \) on the domain \([0, 1]\), plotting can help estimate where the function reaches its peak and trough.

By inspecting a graph, we identify possible extrema by looking for peaks (high points) or troughs (low points) in the plotted curve. Estimation from the graph provides an initial idea of the location of these points, which calculus can then refine. Here, the graph suggested a maximum around \( x = 0.5 \), which was later confirmed through derivative calculations.

• Graphing provides a visual approach to understanding function behavior.
• It offers a preliminary estimate of critical points, subject to verification by calculus.

Derivative Calculation

Calculating the derivative is crucial for understanding how a function changes. In our case, the derivative \( f'(x) \) was found using both the chain rule for derivatives of composite functions and the product rule for derivatives of products of functions.

The process led to the derivative expression:\[ f'(x) = \frac{x(1-2x)}{2\sqrt{x-x^2}} + \sqrt{x-x^2}. \]Setting this derivative to zero allowed identifying critical points. The calculations confirmed that \( x = 0.5 \) was a critical point. We also considered critical points at the domain boundaries, \( x = 0 \) and \( x = 1 \).

• The derivative helps determine how the function grows or shrinks at various points.
• Solving \( f'(x) = 0 \) pinpoints possible extrema within the domain.
• Boundary considerations ensure all potential extrema are examined.