Problem 84 - Algebra 2

Question

Find each composition of functions. Simplify your answer. Let \(f(x)=4 x-1 .\) Find \(\frac{f(a+h)-f(a)}{h}, h \neq 0\)

Step-by-Step Solution

Answer

The value of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) is 4.

Step 1: Evaluating Function at \(a + h\)

According to the function definition, \(f(x) = 4x - 1\). The first step in our calculation is to replace \(x\) with \((a+h)\). So \(f(a+h) = 4(a+h) - 1 = 4a + 4h - 1\)

Step 2: Evaluating Function At \(a\)

Next, we evaluate the function at \(a\), which is \(f(a) = 4a - 1\). Now that we have both \(f(a+h)\) and \(f(a)\), we can put these into the difference quotient formula.

Step 3: Computing The Difference Quotient

The difference quotient is given by \(\frac{f(a+h)-f(a)}{h}\). Plugging in our values, we get \(\frac{(4a + 4h - 1)-(4a - 1)}{h} = \frac{4h}{h}\).

Step 4: Simplifying the Difference Quotient

We simplify the expression by canceling out the \(h\)'s in the numerator and denominator. We get 4 as the result.

Key Concepts

Composition of Functions

The idea behind the composition of functions is that you combine one function with another. In simple terms, it's like doing one function and then another on a given value.
For example, if you have two functions, say \( g(x) \) and \( f(x) \), then the composition \( (f \circ g)(x) \) is simply \( f(g(x)) \).
In the original exercise, the task involves evaluating the function \( f(x) = 4x - 1 \) at different input values such as \( a + h \) and \( a \). Though these aren't exactly function compositions in a traditional sense, they follow a similar process where one expression is evaluated through another.

• Step 1 was simply plugging in \( a + h \) into \( f(x) \).
• Step 2 involved substituting \( a \) into the same function.

This sets the stage for aligning these results into the difference quotient formula, which evaluates the rate of change.

Function Evaluation

Function evaluation involves substituting a given value into the function's expression and solving it.
For instance, if we have \( f(x) = 4x - 1 \), and we need to find \( f(a + h) \), we replace \( x \) in \( 4x - 1 \) with \( a + h \).
This transforms our function into \( 4(a + h) - 1 = 4a + 4h - 1 \).

• It's crucial to notice how \( (a + h) \) was multiplied by 4, following the function's rule.
• The same method is applied to find \( f(a) \) by replacing \( x \) with \( a \), resulting in \( 4a - 1 \).

This method of replacing variables in a function with different expressions helps solve various mathematical problems by showing different outcomes.

Simplification

Simplification takes our mathematical expression and makes it as simple as possible. It's about reducing complexities while maintaining mathematical accuracy.
In the original exercise, once the difference quotient formula \( \frac{f(a+h)-f(a)}{h} \) is set up, simplification steps in.

• First, the evaluated forms, \( f(a+h) = 4a + 4h - 1 \) and \( f(a) = 4a - 1 \), are substituted into the difference quotient, resulting in \( \frac{(4a + 4h - 1) - (4a - 1)}{h} \).
• The simplification occurs when common terms are canceled out, leading to \( \frac{4h}{h} \).
• Finally, cancelling the \( h \)s gives us the outcome of 4.

Simplification removes extra "weight" in expressions, offering a clearer and often insightful perspective on the problem.

Algebraic Expressions

Understanding algebraic expressions is key to solving many math problems, including function evaluation.
An algebraic expression like \( 4x - 1 \) uses variables and constants to represent numbers logically.
In the steps given, we saw how \( 4(a + h) - 1 \) expanded due to distribution of the 4 over the \( (a + h) \), changing it to \( 4a + 4h - 1 \).

• Algebraic manipulation enables alterations, such as moving terms around and simplifying them, which is essential in evaluating expressions efficiently.
• Operators such as addition, subtraction, multiplication, and division are often used to connect different parts of the expression and maintain its balance.

Mastering how to work with these expressions empowers you to engage confidently with more complex mathematical challenges.