Problem 65 - A Graphical Approach to Precalculus with Limits

Question

For the given \(f(x)\) find (a) \(f(a),\) (b) \(f(b+1),\) and (c) \(f(3 x)\). $$f(x)=5 x$$

Step-by-Step Solution

Answer

(a) \( f(a) = 5a \), (b) \( f(b+1) = 5b + 5 \), (c) \( f(3x) = 15x \).

Step 1: Understand the function

The given function is a linear function, denoted as \( f(x) = 5x \). This means that for any input \( x \), the output is the input multiplied by 5.

Step 2: Evaluate \( f(a) \)

To find \( f(a) \), substitute \( a \) for \( x \) in the function: \( f(a) = 5a \). This indicates that when the input is \( a \), the output is 5 times \( a \).

Step 3: Evaluate \( f(b+1) \)

To find \( f(b+1) \), substitute \( b+1 \) for \( x \) in the function: \( f(b+1) = 5(b+1) = 5b + 5 \). Therefore, for an input of \( b+1 \), the output is \( 5b + 5 \).

Step 4: Evaluate \( f(3x) \)

To find \( f(3x) \), substitute \( 3x \) for \( x \) in the function: \( f(3x) = 5(3x) = 15x \). This shows that for an input of \( 3x \), the output is \( 15x \).

Key Concepts

Function Evaluation

Function evaluation is the process of determining the output of a function for a given input. In the context of linear functions, this involves substituting a specific value into the function equation to find the resultant value. For example, given the linear function \( f(x) = 5x \), and we want to evaluate \( f(a) \), we substitute \( a \) into the function equation.

• Start with the function equation: \( f(x) = 5x \)
• Replace \( x \) with the given input, here \( a \)
• The evaluated function becomes \( f(a) = 5a \)

This process shows how the input \( a \) is multiplied by 5 to obtain the output. Function evaluation is fundamental because it allows you to determine how the function behaves with different inputs. It is essential to understand that function evaluation simply means finding the value of a function at a certain point, which is a basic aspect of working with functions.

Substitution in Functions

Substituting in functions involves replacing the variable in the function equation with a specific expression. This technique is crucial for rewriting and finding values of functions for different scenarios. Let's explore this with a linear function like \( f(x) = 5x \).When you need to evaluate the function at a particular point or expression, such as \( f(b+1) \), you follow these straightforward steps:

• Take the original function: \( f(x) = 5x \)
• Substitute \( b+1 \) for \( x \)
• The new function becomes \( f(b+1) = 5(b+1) \, = \, 5b + 5 \)

Substitution allows flexibility in evaluating functions for composite arguments like \( b+1 \). It’s a transitioning step where you modify the function to suit the given input. This is a basic yet very powerful tool, especially since it's applicable in many areas like transforming functions and solving equations.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make them easier to understand or solve. When dealing with linear functions, this often means reorganizing terms or distributing, as seen in evaluating expressions like \( f(b+1) \) or \( f(3x) \) for the function \( f(x) = 5x \).Here's how manipulation plays out when evaluating \( f(3x) \):

• Begin with the substitution: replace \( x \) with \( 3x \) to get \( f(3x) = 5(3x) \).
• Perform the multiplication: \( 5 \times 3 \) to get \( 15x \).
• The function simplifies to \( f(3x) = 15x \).

Algebraic manipulation is crucial in processes like distributing, combining like terms, and simplifying. It makes it possible to solve equations more efficiently by transforming complex expressions into simpler forms. In linear functions, this manipulation helps illustrate how different inputs affect the output in a clear and concise way.