Problem 71 - Intermediate Algebra

Question

Let \(f(x)=\frac{1}{x}\) and \(g(x)=\frac{1}{x^{2}} .\) Find each of the following. $$ (g \circ f)(8 x) $$

Step-by-Step Solution

Answer

The result of \((g \circ f)(8x)\) is \(64x^2\).

Step 1: Understand Composition of Functions

When we have two functions, say \(f(x)\) and \(g(x)\), the composition \((g \circ f)(x)\) means we first apply \(f\) to \(x\), and then apply \(g\) to the result of \(f(x)\). So mathematically, \((g \circ f)(x) = g(f(x))\).

Step 2: Substitute f(x) into g

Let's apply this rule to find \((g \circ f)(8x)\):First, calculate \(f(8x)\). Since \(f(x) = \frac{1}{x}\), we have:\[ f(8x) = \frac{1}{8x} \]Next, substitute \(f(8x)\) into \(g(x)\), where \(g(x) = \frac{1}{x^2}\). This gives us:\[ g(f(8x)) = g\left(\frac{1}{8x}\right) \]

Step 3: Compute g(f(8x))

Now we need to compute \(g\left(\frac{1}{8x}\right)\). Since \(g(x) = \frac{1}{x^2}\), replace \(x\) with \(\frac{1}{8x}\):\[ g\left(\frac{1}{8x}\right) = \frac{1}{\left(\frac{1}{8x}\right)^2} \]Simplify the expression inside the square by squaring the fraction:\[ \left(\frac{1}{8x}\right)^2 = \frac{1}{(8x)^2} = \frac{1}{64x^2} \]Thus, the function becomes:\[ \frac{1}{\left(\frac{1}{64x^2}\right)} = 64x^2 \]

Step 4: Conclusion

Hence, we conclude that:\[ (g \circ f)(8x) = 64x^2 \]

Key Concepts

Composition of Functions

When dealing with the composition of functions, the main idea is to combine two functions so that the output of one function becomes the input of another function. This process allows you to create new functions and explore how different operations interact with one another.
In notation, if you have two functions—let's say, function \( f \) and function \( g \)—the composition is denoted as \((g \circ f)(x)\). This means "g of f of x," or to put it simply, you first apply \( f \) to \( x \), and then you take the result of \( f(x) \) and pass it into \( g \).

• First Step: Evaluate \( f(x) \).
• Second Step: Use the result from \( f(x) \) as the input for \( g(x) \).

This is a key concept in higher mathematics as it allows for more complex equations and solutions, making it easier to solve real-world problems or engage in abstract mathematical investigations.

Intermediate Algebra

Intermediate algebra is a level of mathematics that focuses on expanding basic algebraic principles. In intermediate algebra, students will encounter concepts like function operations, polynomials, rational expressions, and equations.
At this level, you'll learn to manipulate and transform different algebraic expressions through various operations including composition of functions. Understanding these operations is crucial as they form the backbone of higher-level math, such as calculus.
The key elements of intermediate algebra include:

• Operations with Functions: Learn how to add, subtract, multiply, divide, and compose functions.
• Rational Expressions: Work with ratios of polynomials, similar to how fractions are managed in basic arithmetic.
• Solving Equations: Attack increasingly complex equations, including those involving multiple operations.

While this level might seem challenging at first, mastering it opens the door to even more exciting mathematical topics.

Function Operations

Function operations allow you to perform calculations with functions, similar to the numerical operations you are already familiar with. With functions, you not only perform the basic operations of addition, subtraction, multiplication, and division, but you can also compose them.
When dealing with function operations:

• Addition: If \( f(x) \) and \( g(x) \) are two functions, then \( (f + g)(x) = f(x) + g(x) \).
• Subtraction: The operation is \( (f - g)(x) = f(x) - g(x) \).
• Multiplication: This provides \( (f \cdot g)(x) = f(x) \cdot g(x) \).
• Division: Given by \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \), as long as \( g(x) eq 0 \).
• Composition: As explained earlier, \( (g \circ f)(x) = g(f(x)) \).

Understanding how these operations work allows you to manipulate and solve more complex problems that involve multiple functions. This skill becomes especially useful in various areas of mathematics, such as calculus, where function operations play a central role.