Problem 61 - Elementary and Intermediate Algebra

Question

Find \(h(5)\) and \(h(-2) .\) See Example 4. $$ h(x)=\frac{1}{x+3} $$

Step-by-Step Solution

Answer

\( h(5) = \frac{1}{8} \) and \( h(-2) = 1 \).

Step 1: Introducing the Function

We are given the function \( h(x) = \frac{1}{x+3} \). Our task is to evaluate this function at specific points: \( x = 5 \) and \( x = -2 \).

Step 1: Evaluate \( h(5) \)

Substitute \( x = 5 \) into the function: \( h(5) = \frac{1}{5 + 3} = \frac{1}{8} \). Thus, \( h(5) = \frac{1}{8} \).

Step 2: Evaluate \( h(-2) \)

Substitute \( x = -2 \) into the function: \( h(-2) = \frac{1}{-2 + 3} = \frac{1}{1} = 1 \). Hence, \( h(-2) = 1 \).

Key Concepts

Understanding Rational Functions

Rational functions are a type of function where one polynomial is divided by another polynomial. They take the form \( f(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials. It is important to note that the denominator, \( Q(x) \), cannot be zero because division by zero is undefined. This will cause the function to have "holes" or vertical asymptotes where the function is not defined at those points.

In the context of our original exercise, the function given is \( h(x) = \frac{1}{x+3} \), which is a simple rational function. Here, the numerator is a constant 1, and the denominator is \( x + 3 \). The function \( h(x) \) is undefined when \( x = -3 \), as plugging \( x = -3 \) into the denominator would make it zero. By analyzing these aspects, we can better understand how to approach evaluating such functions at specific points in their domain.

Exploring the Substitution Method

The substitution method is a straightforward approach for evaluating functions at specific points. It involves replacing a variable in the function with a given number, allowing us to compute the value of the function at that specific point.

Let's apply this method to the function \( h(x) = \frac{1}{x+3} \).

• Substituting \( x = 5 \) into the function: \( h(5) = \frac{1}{5+3} = \frac{1}{8} \).
• Substituting \( x = -2 \) into the function: \( h(-2) = \frac{1}{-2+3} = \frac{1}{1} = 1 \).

Both steps illustrate that the function value is found by replacing \( x \) with the specified value and simplifying the resulting expression. The key is to correctly substitute and simplify without algebraic errors, thus yielding the correct function output.

Demystifying Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. They form the backbone of algebra, serving as representations of relationships between quantities. Understanding how to manipulate these expressions allows us to solve equations, evaluate functions, and model real-world scenarios.

In the given exercise, the expression \( \frac{1}{x+3} \) exemplifies an algebraic expression where the variable \( x \) determines the value of the entire expression. When evaluating algebraic expressions, like in our substitution method for \( h(x) \), we focus on

• Simplifying expressions by performing operations.
• Substituting values for variables to calculate specific outputs.
• Ensuring the denominator is never zero, to avoid undefined expressions.

Algebraic expressions provide a systematic approach to performing calculations needed in various applications of mathematics, such as problem-solving and analysis.