Problem 63 - Calculus Early Transcendentals

Question

Use a graphing utility to make rough estimates of the intervals on which \(f^{\prime}(x)>0,\) and then find those intervals exactly by differentiating. $$ f(x)=x-\frac{1}{x} $$

Step-by-Step Solution

Answer

The intervals are \((-\infty, 0)\cup(0,\infty)\).

Step 1: Differentiate the Function

Given the function \( f(x) = x - \frac{1}{x} \). We need to find the derivative \( f'(x) \). Using standard differentiation rules: the derivative of \( x \) is \( 1 \) and the derivative of \( -\frac{1}{x} \) is \( \frac{1}{x^2} \). Thus, \( f'(x) = 1 + \frac{1}{x^2} \).

Step 2: Set the Derivative Greater Than Zero

We need to find the intervals where \( f'(x) > 0 \). Thus, we set the inequality: \[ 1 + \frac{1}{x^2} > 0 \].

Step 3: Interpret the Inequality

The inequality \( 1 + \frac{1}{x^2} > 0 \) is always true because \( \frac{1}{x^2} \) is always non-negative (and positive when \( x eq 0 \)), and adding \( 1 \) ensures that the value is greater than \( 0 \). Thus, \( f'(x) > 0 \) for all \( x eq 0 \).

Step 4: Identify the Exact Interval

Given that \( f'(x) = 1 + \frac{1}{x^2} \) is greater than zero for all \( x eq 0 \), the exact intervals where \( f'(x) > 0 \) are \( (-\infty, 0) \cup (0, \infty) \).

Key Concepts

Differentiation

Differentiation is a core concept in calculus, allowing us to find the rate at which a function is changing at any point. When we differentiate a function, we are essentially finding its derivative. Derivatives help us understand how functions behave and where they increase or decrease.
In the given exercise, the function is \( f(x) = x - \frac{1}{x} \). To differentiate this function, we apply standard rules of differentiation:

• The derivative of \( x \) is \( 1 \).
• The derivative of \(-\frac{1}{x}\) can be rewritten as \(-x^{-1}\). Using the power rule, its derivative is \(\frac{1}{x^2}\).

Thus, the derivative of \( f(x) \) is \( f'(x) = 1 + \frac{1}{x^2} \). This derivative provides us with crucial information about the slope of the tangent line at any point on the graph of the original function.

Graphing utility

A graphing utility is a powerful tool for visualizing mathematical concepts. It allows students to plot functions and their derivatives to make interpretations based on visual data. Graphing can help in estimating where a function is increasing or decreasing, based on the slope of the tangent line at different points on the graph.
To solve the exercise, using a graphing utility helps to:

• Plot the function \( f(x) = x - \frac{1}{x} \).
• Estimate the intervals where its derivative \( f'(x) \) is positive.
• Visualize the behavior of the function around critical points such as \( x = 0 \).

By employing a graphing utility before differentiating, you can gain an intuition about the function's behavior, preparing you to calculate more precise intervals analytically.

Inequalities

Inequalities are vital in calculus to determine the behavior of functions over different intervals. In this exercise, we derive the inequality \( 1 + \frac{1}{x^2} > 0 \) from the derivative \( f'(x) = 1 + \frac{1}{x^2} \).
Solving this inequality involves basic principles:

• Recognize that \( \frac{1}{x^2} \) is always non-negative because squaring any real number, except zero, gives a positive result.
• Addition of \( 1 \) means the combined result will always be greater than zero, no matter the value of \( x \) (except at \( x = 0 \)).

Thus, the inequality \( 1 + \frac{1}{x^2} > 0 \) simplifies to say that \( f'(x) \) is positive for all values of \( x \) except zero, providing us with the needed intervals for further analysis.

Intervals

Intervals are sections of the number line where we evaluate particular behaviors of functions, such as increasing or decreasing trends. When we analyze the derivative \( f'(x) = 1 + \frac{1}{x^2} \), it tells us that outside of \( x = 0 \), the function is always increasing.
Here's how we establish the intervals for \( f'(x) > 0 \):

• The function is undefined at \( x = 0 \), thus we exclude this point. No information about slope is provided there.
• For any \( x eq 0 \), \( f'(x) = 1 + \frac{1}{x^2} \) ensures a positive result.

Hence, the exact intervals where \( f'(x) \) remains positive are identified as \( (-\infty, 0) \cup (0, \infty) \). These intervals express where the original function is increasing, excluding the undefined point at zero.