Problem 77 - Calculus Early Transcendentals

Question

In each part, find \(k\) so that \(f\) has a relative extremum at the point where \(x=3 .\) $$ \text { (a) } f(x)=x^{2}+\frac{k}{x} \quad \text { (b) } f(x)=\frac{x}{x^{2}+k} $$

Step-by-Step Solution

Answer

Part (a): \( k = 54 \); Part (b): \( k = 9 \).

Step 1: Differentiate the Function

First, we'll find the first derivative of function \( f(x) = x^2 + \frac{k}{x} \). We need this to locate any critical points where relative extrema could occur. The derivative is \( f'(x) = 2x - \frac{k}{x^2} \).

Step 2: Determine the Critical Point

The critical points occur when the derivative is zero. Set \( f'(x) = 0 \): \( 2x - \frac{k}{x^2} = 0 \). Solve for \( k \) given the condition \( x = 3 \) yields \( 2 \times 3 - \frac{k}{3^2} = 0 \).

Step 3: Solve for k in Part (a)

Substitute \( x = 3 \) into the equation from Step 2: \( 6 - \frac{k}{9} = 0 \). Solve this equation to find \( k \). Multiply through by 9 to obtain \( k = 54 \).

Step 4: Differentiate the Second Function

Now, find the first derivative of the second function \( f(x) = \frac{x}{x^2 + k} \). Use the quotient rule: \( f'(x) = \frac{(x^2 + k)(1) - x(2x)}{(x^2 + k)^2} = \frac{k - x^2}{(x^2 + k)^2} \).

Step 5: Determine the Critical Point for Part (b)

Set the derivative equal to zero: \( \frac{k - x^2}{(x^2 + k)^2} = 0 \). This gives \( k - x^2 = 0 \). Substitute \( x = 3 \) into this equation to find \( k = x^2 = 9 \).

Step 6: Final Step: Verify Solutions

The step confirms that both values for \( k \) provide critical points at \( x=3 \). Part (a) yields \( k=54 \) and part (b) yields \( k=9 \) as solutions.

Key Concepts

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how the function changes as its input changes. It is a fundamental concept in calculus, allowing for the analysis of rates of change and the behavior of functions. To differentiate a function, you apply rules such as the power rule, product rule, and quotient rule.

In the given exercise, we start by differentiating the functions to find the critical points. For the function \( f(x) = x^2 + \frac{k}{x} \), the process requires using the power rule and the rule for differentiating a term of the form \( \frac{k}{x} \), resulting in \( f'(x) = 2x - \frac{k}{x^2} \).

For the second function, \( f(x) = \frac{x}{x^2+k} \), the quotient rule is used: \( f'(x) = \frac{(x^2 + k)(1) - x(2x)}{(x^2 + k)^2} \). The quotient rule helps differentiate functions expressed as one function divided by another, ensuring an accurate derivative calculation.

Critical Points

Critical points of a function occur where the derivative of the function is zero or undefined. These points are important because they can indicate where a function changes direction, which might be a maximum, minimum, or a point of inflection.

In this exercise, we locate the critical points by setting the derivative to zero. For part (a), setting \( f'(x) = 2x - \frac{k}{x^2} = 0 \) and solving for \( x = 3 \) helps us find \( k \). In part (b), setting \( f'(x) = \frac{k - x^2}{(x^2 + k)^2} = 0 \) and again substituting \( x = 3 \) allows us to determine \( k \).

When the derivative equals zero, this typically identifies potential relative extrema. By finding the correct \( k \), we ensure a critical point at \( x = 3 \) for both functions provided in the exercise.

Relative Extrema

Relative extrema refer to the relative maximum or minimum points of a function. These are points where the function changes direction from increasing to decreasing or vice versa, but they might not be the highest or lowest points on the entire graph.

To identify a relative extremum at \( x = 3 \), we use the critical points found through differentiation. In the exercise, by adjusting the parameter \( k \), we confirm that the functions have maxima or minima at \( x = 3 \).

For function (a), with the determined \( k = 54 \), and for function (b), with \( k = 9 \), these \( k \) values make \( x = 3 \) a critical point where potential relative extrema could occur. Further analysis, such as the second derivative test, could confirm whether these are indeed maxima or minima.