Problem 32 - Calculus Early Transcendentals

Question

Use a graphing utility to estimate the absolute maximum and minimum values of \(f \), if any, on the stated interval, and then use calculus methods to find the exact values. \(f(x)=\frac{x}{x^{2}+2} ;[-1,4]\)

Step-by-Step Solution

Answer

Absolute maximum: \(0.354\) at \(x = \sqrt{2}\); absolute minimum: \(-0.354\) at \(x = -\sqrt{2}\).

Step 1: Identify Critical Points

To find the critical points of the function \( f(x) = \frac{x}{x^2 + 2} \) in the interval \([-1, 4]\), we start by finding the derivative \( f'(x) \) using the quotient rule. The derivative is \( f'(x) = \frac{1 \cdot (x^2 + 2) - x \cdot 2x}{(x^2 + 2)^2} = \frac{-x^2 + 2}{(x^2 + 2)^2} \). Set the numerator equal to zero: \(-x^2 + 2 = 0\). Solve for \(x\), giving \(x = \pm \sqrt{2}\). Additionally, check the interval endpoints, \(x = -1\) and \(x = 4\).

Step 2: Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical points and endpoints. For \( x = -1 \), we have \( f(-1) = \frac{-1}{(-1)^2 + 2} = -\frac{1}{3} \). For \( x = \sqrt{2} \), \( f(\sqrt{2}) = \frac{\sqrt{2}}{(\sqrt{2})^2 + 2} = \frac{\sqrt{2}}{4} \). For \( x = -\sqrt{2} \), \( f(-\sqrt{2}) = \frac{-\sqrt{2}}{4} \). For \( x = 4 \), \( f(4) = \frac{4}{16 + 2} = \frac{2}{9} \).

Step 3: Compare Function Values

Compare the calculated function values: \(-\frac{1}{3} \), \( \frac{\sqrt{2}}{4} \), \( -\frac{\sqrt{2}}{4} \), and \( \frac{2}{9} \). Note that \(-\frac{1}{3} \) is approximately \(-0.333\), \( \frac{\sqrt{2}}{4} \) is approximately \(0.354\), \(-\frac{\sqrt{2}}{4} \) is approximately \(-0.354\), and \( \frac{2}{9} \) is approximately \(0.222\). Thus, the absolute maximum is approximately \(0.354\) at \(x = \sqrt{2}\) and the absolute minimum is approximately \(-0.354\) at \(x = -\sqrt{2}\).

Step 4: Confirm with Graphing Utility

Use a graphing utility to visualize \(f(x)\) over the interval \([-1, 4]\). Confirm that the graph shows peaks and valleys corresponding to our calculated values. The highest point on the graph in this interval occurs near \(x = \sqrt{2}\) and the lowest point near \(x = -\sqrt{2}\), confirming the estimated maximum and minimum.

Key Concepts

Critical Points

When studying calculus, critical points play a crucial role in understanding how a function behaves across different intervals. Critical points are locations on the graph of a function where the first derivative is zero or undefined. These points can indicate potential maxima, minima, or inflection points.

To find these points, you start by taking the derivative of the function and setting it equal to zero, as shown in the exercise. For example, the function given, \(f(x) = \frac{x}{x^2 + 2}\), has its derivative calculated as \(f'(x) = \frac{-x^2 + 2}{(x^2 + 2)^2}\).

The roots of the equation \(-x^2 + 2 = 0\) provide the critical points \(x = \pm \sqrt{2}\). It is important also to consider the endpoints of the interval, \([-1, 4]\), as they could represent critical points for the specific bounded range of the function.

Absolute Maximum

Finding the absolute maximum of a function on a closed interval involves evaluating the function at its critical points and endpoints. An absolute maximum is the highest value a function reaches within a given domain.

To establish this, calculate the function at critical points and endpoints as illustrated in the problem:

• \(f(-1) = -\frac{1}{3}\)
• \(f(\sqrt{2}) = \frac{\sqrt{2}}{4}\)
• \(f(4) = \frac{2}{9}\)

After these evaluations, the largest value among them is approximately \(\frac{\sqrt{2}}{4} \approx 0.354\), making it the absolute maximum on the interval at \(x = \sqrt{2}\).

Recognizing these values with a graphing utility aids in confirming these points visually, ensuring the calculations correlate with the actual behavior of the function's graph in the specified interval.

Absolute Minimum

While searching for the absolute minimum of a function is similar to finding the maximum, it seeks the smallest value instead. The absolute minimum is the lowest point of a function's output within a defined domain.

Using the same evaluations as described, the absolute minimum value can be found by checking:

• \(f(-\sqrt{2}) = -\frac{\sqrt{2}}{4}\)
• \(f(-1) = -\frac{1}{3}\)
• \(f(4) = \frac{2}{9}\)

Compared to other values, \(-\frac{\sqrt{2}}{4}\), which is approximately \(-0.354\), represents the absolute minimum at \(x = -\sqrt{2}\).

Through graphing, this minimal point should visually appear as the lowest dip on the function in the defined interval, further substantiating the role of calculus in defining function behavior through numerical verification and graphical analysis.

Quotient Rule

The Quotient Rule is a fundamental derivative tool in calculus, especially useful for functions represented as the division of two functions. It allows for effective calculation of derivatives in such cases.

The rule states that if \(y = \frac{u}{v}\), then the derivative \(y'\) is given by:
\[ y' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Here, \(u\) is the numerator and \(v\) is the denominator. In the context of the provided function, \(u = x\) and \(v = x^2 + 2\), and their derivatives, \(u'\) and \(v'\), are critical for solving the problem.

This method was applied to find the derivative \(f'(x)\), which played a vital role in understanding where the function had critical points. Becoming proficient with the Quotient Rule empowers students to handle a wide array of functions and interpret analyses of real-world models.