Problem 38 - Thomas Calculus

Question

Define \(h(2)\) in a way that extends \(h(t)=\left(t^{2}+3 t-10\right) /(t-2)\) to be continuous at \(t=2\) .

Step-by-Step Solution

Answer

Define \( h(2) = 7 \) to make the function continuous at \( t=2 \).

Step 1: Understand Continuity

To make a function continuous at a point, the function value at that point must equal the limit of the function as it approaches that point. We need to find the limit of \( h(t) \) as \( t \rightarrow 2 \) and define \( h(2) \) to be this limit.

Step 2: Simplify the Function

To find the limit, simplify the expression \( \frac{t^2 + 3t - 10}{t-2} \). Factor the numerator to find potential cancellations with the denominator.

Step 3: Factor and Cancel

The numerator \( t^2 + 3t - 10 \) can be factored as \( (t-2)(t+5) \). The function then becomes \( \frac{(t-2)(t+5)}{t-2} \). Cancel the \( (t-2) \) terms, leaving \( t+5 \).

Step 4: Find the Limit

Now, find the limit as \( t \rightarrow 2 \) of the simplified function, \( t+5 \). This is simply \( 2 + 5 = 7 \).

Step 5: Define h(2)

To make \( h(t) \) continuous at \( t=2 \), define \( h(2) = 7 \), since this is the limit of \( h(t) \) as \( t \) approaches 2.

Key Concepts

Understanding Limits of Functions

When dealing with continuity, one of the key concepts is the limit of a function. A limit essentially measures the value that a function approaches as the input approaches a certain point. If we say "the limit of \( h(t) \) as \( t \rightarrow 2 \) equals 7," it means that as \( t \) gets closer and closer to 2, the value of \( h(t) \) gets closer and closer to 7.
This helps us understand the behavior of functions around specific points.
To determine this behavior, you don't actually need to reach the point—just close enough to it.

• In our case, since the function was originally undefined at \( t = 2 \), evaluating the limit gives us a way to "fill in the gap."
• This allows us to make the function continuous by defining the function's value at 2 as this limit.

Limits are a fundamental part of calculus and are essential when working on the continuity of functions.

Factoring Polynomials in Limits

Factoring polynomials is an important skill, especially in calculus, because it helps simplify expressions, particularly when you're dealing with limits. When a function includes a polynomial divided by another expression, as we saw in the example, simplifying is crucial.
The given function \( \frac{t^2 + 3t - 10}{t-2} \) required factoring the numerator \( t^2 + 3t - 10 \).

• Factoring involves rewriting the polynomial as a product of its roots: here, \( (t-2)(t+5) \).
• This step is crucial for cancelling common terms in the fraction and simplifying the expression.

After cancelling the \( (t-2) \) terms, we are left with \( t+5 \), which is much easier to work with for finding limits. Understanding how polynomials factor directly impacts your ability to manipulate and simplify expressions, revealing more about the function's behavior around certain points.

Rational Functions and Continuity

Rational functions, like \( h(t) = \frac{t^2 + 3t - 10}{t-2} \), are fractions where both the numerator and the denominator are polynomials.
Continuity in rational functions often depends on whether or not there are common factors between the numerator and denominator. If these factors exist and cancel, like in this function, it's possible to "repair" parts of the function that are initially undefined at certain points.

• Cancelling the common factor \( (t-2) \) left \( t+5 \), which we can evaluate directly for any \( t \), including \( t = 2 \).
• Defining \( h(2) \) to be the limit allows the rational function to be extended and continuous at \( t = 2 \). This means the function behaves normally without any jumps or holes at this point.

This concept is integral to making sense of how rational functions behave, and it's key to ensuring the smooth functionality of a function across its domain.