Problem 37 - Thomas Calculus

Question

Define \(g(3)\) in a way that extends \(g(x)=\left(x^{2}-9\right) /(x-3)\) to be continuous at \(x=3 .\)

Step-by-Step Solution

Answer

Define \( g(3) = 6 \) to make \( g(x) \) continuous at \( x = 3 \).

Step 1: Understand the Problem

We are given the function \( g(x) = \frac{x^2 - 9}{x - 3} \) and need to define \( g(3) \) so that the function becomes continuous at \( x = 3 \). This means the limit of \( g(x) \) as \( x \) approaches 3 needs to equal \( g(3) \).

Step 2: Simplify the Expression

Notice that \( x^2 - 9 \) is a difference of squares and can be factored: \( x^2 - 9 = (x-3)(x+3) \). Substitute this back into the function: \[ g(x) = \frac{(x-3)(x+3)}{x-3}. \]

Step 3: Cancel the Common Factor

Since \( x eq 3 \) for the original function (the domain excludes \( x = 3 \) due to division by zero), we can cancel \( x-3 \) in the numerator and denominator:\[g(x) = x + 3 \] for \( x eq 3 \).

Step 4: Compute the Limit

Now let's find the limit of \( g(x) \) as \( x \to 3 \). Use the simplified version to find the limit:\[\lim_{{x \to 3}} (x + 3) = 3 + 3 = 6. \]

Step 5: Define \( g(3) \) for Continuity

To make \( g(x) \) continuous at \( x = 3 \), define \( g(3) \) as the value of the limit:\( g(3) = 6 \).

Step 6: Verify Continuity

Now \( g(x) \) is continuous at \( x = 3 \) because \( \lim_{{x \to 3}} g(x) = g(3) = 6 \).

Key Concepts

Understanding Limits

Limits are a foundational concept in calculus and are essential for understanding continuity of functions. A limit essence is to describe the behavior of a function as the input approaches a certain point. When we say the limit of a function \( g(x) \) as \( x \to c \) is \( L \), it means as \( x \) gets closer to \( c \), \( g(x) \) gets closer to \( L \).
In our exercise, we need to find the limit of \( g(x) = \frac{x^2 - 9}{x - 3} \) as \( x \to 3 \). The trick is simplifying the expression first, which we did through factoring. Once simplified, we see the behavior of the function is just a straight line, \( x + 3 \), when \( x eq 3 \).
To ensure the function is continuous at \( x = 3 \), the limit we calculated \( \lim_{{x \to 3}} (x + 3) = 6 \) must equal the function value at that point. By defining \( g(3) = 6 \), we extend the function smoothly over \( x = 3 \), achieving continuity.

Factoring in Mathematics

Factoring is a powerful technique in mathematics, essential for simplifying complex expressions and solving equations, particularly useful in finding limits. When faced with complex expressions like \( x^2 - 9 \), recognizing it as a difference of squares is crucial. The difference of squares is expressed as \((a^2-b^2) = (a-b)(a+b)\).
In our exercise, \( x^2 - 9 = (x-3)(x+3) \). By factoring, we can cancel out the \( x - 3 \) from the denominator in \( g(x) = \frac{(x-3)(x+3)}{x-3} \), thus simplifying the function to \( g(x) = x + 3 \) for \( x eq 3 \).
This simplification was crucial for calculating the limit at \( x = 3 \), which could not have been directly computed from the original expression due to division by zero. Thus, factoring not only aids in simplification but also enables further analysis, like limit computation.

Working with Piecewise Functions

Piecewise functions are defined by different expressions over different intervals of the domain. They allow for functions to be flexible and cater to different conditions.
In making \( g(x) \) continuous at \( x = 3 \), it becomes a piecewise function. Originally undefined at \( x = 3 \), the provided solution fills the gap by defining \( g(3) = 6 \). Consequently, \( g(x) \) can be expressed as:

• \( g(x) = x + 3 \) when \( x eq 3 \)
• \( g(x) = 6 \) when \( x = 3 \)

Defining functions piecewise ensures smoothness and continuity across intervals, addressing behaviors or sudden changes like filling a point of discontinuity, as we did here to satisfy continuity at a specific point.