Problem 5 - Calculus with Concepts in Calculus

Question

Find all critical numbers of the given function. $$ g(x)=x+1 / x $$

Step-by-Step Solution

Answer

The critical numbers are \( x = 1 \) and \( x = -1 \).

Step 1: Find the Derivative

To find the critical numbers, we first need to determine the derivative of the function. The given function is:\[ g(x) = x + \frac{1}{x} \]Using the power rule and the derivative of a reciprocal, the derivative is:\[ g'(x) = 1 - \frac{1}{x^2} \]

Step 2: Set the Derivative to Zero

Set the derivative equal to zero to find the critical points:\[ 1 - \frac{1}{x^2} = 0 \]Solve for \(x\):\[ 1 = \frac{1}{x^2} \]This means \( x^2 = 1 \).

Step 3: Solve for x

Solve the equation \( x^2 = 1 \) to find the values of \(x\):\[ x = \pm 1 \]Thus, the potential critical numbers are \( x = 1 \) and \( x = -1 \).

Step 4: Determine the Domain of the Original Function

Since the original function \( g(x) = x + \frac{1}{x} \) includes a term \( \frac{1}{x} \), the function is not defined for \( x = 0 \).Thus, \( x = 0 \) is excluded from the critical numbers.

Step 5: Identify the Valid Critical Numbers

The critical numbers are the values from Step 3 that lie in the domain of \( g(x) \), which excludes \( x = 0 \). Thus, the critical numbers of the function are \( x = 1 \) and \( x = -1 \).

Key Concepts

The Derivative: Finding the Rate of Change

To find critical numbers of a function, determining the derivative is a crucial first step. The derivative measures how a function changes at each point. For the function \( g(x) = x + \frac{1}{x} \), finding the derivative highlights where changes stop or turn around.

Calculating derivatives involves different methods depending on the function. In this exercise, knowing how to differentiate a sum of functions, like \( x \) and \( \frac{1}{x} \), is essential.

• The derivative of \( x \) is straightforward, resulting in 1 because each unit increase in \( x \) results in a one-unit increase in \( g(x) \).
• For the term \( \frac{1}{x} \), its derivative relies on special derivative formulas, like the reciprocal derivative, which we will cover.

Practicing derivative calculations enhances your ability to determine critical numbers, helping you understand function behavior.

Unraveling the Power Rule

The power rule is a fundamental tool for taking derivatives, widely applicable when a function includes terms like \( x^n \). The rule simplifies finding derivatives for any power of \( x \).

To apply the power rule:

• Multiply the coefficient by the power.
• Reduce the power by one.

For example, if we had \( x^1 \) in our function, the power rule confirms its derivative is 1.

In this exercise, differentiating \( x \) as part of \( g(x) = x + \frac{1}{x} \) is a direct application of the power rule, showcasing its simplicity and effectiveness.

Exploring the Function Domain

Understanding the domain of a function is crucial when identifying critical numbers. The domain is the set of all possible inputs \( x \) for a function, affecting its critical points.

For \( g(x) = x + \frac{1}{x} \), notice the term \( \frac{1}{x} \). Here, \( x = 0 \) is not allowed because division by zero is undefined.

As a result, the domain of \( g(x) \) excludes \( x = 0 \). When considering critical numbers, it's important to only include those within the domain. Thus, even if solving for critical numbers gives us potential values, they must be checked against the function's domain.

Understanding Reciprocal Derivatives

The reciprocal derivative refers to the process of deriving a function like \( \frac{1}{x} \). This is essential when the function involves inverse or reciprocal terms.

For \( \frac{1}{x} \):

• Recognize it can be expressed as \( x^{-1} \).
• Apply the power rule: start by bringing down the power (e.g., -1) and reduce the power by one, leading to \(-x^{-2}\).

Hence, the derivative of \( \frac{1}{x} \) becomes \( -\frac{1}{x^2} \).

This derivative tells us how the reciprocal function’s rate of change behaves, key in identifying critical numbers and overall understanding the role of reciprocal functions in calculus.