Problem 40 - Algebra 2

Question

Let \(f(x)=x^{2}\) and \(g(x)=x-3 .\) Find each value or expression. $$ (g \circ f)(c) $$

Step-by-Step Solution

Answer

The value of expression \((g \circ f)(c)\) is \((c^{2})-3\).

Step 1: Compose the functions

Firstly, we need to find \(g \circ f(x)\). According to the definition of function composition, we have to substitute \(f(x)\) into \(g(x)\). So, \(g \circ f(x) = g(f(x))\).

Step 2: Substitute \(f(x)\) into \(g(x)\)

As given, the functions are \(f(x)=x^{2}\) and \(g(x)=x-3 \). So, \(g(f(x)) = g(x^{2}) = (x^{2})-3\). This gives us the composite function \(g \circ f(x)\).

Step 3: Find \(g \circ f(c)\)

We are finding \(g \circ f\) at a general value \(c\). So substitute \(x\) with \(c\) in \(g \circ f(x)\) i.e., if \(g \circ f(x) = (x^{2})-3\), then \(g \circ f(c) = (c^{2})-3\).

Key Concepts

Composite Function

When dealing with functions like baby puzzles, think about the Composite Function as layering one onto another. The composite function, symbolized as \( (g \circ f)(x) \), means you're first applying one function, and then immediately applying another. It's like taking a detour on your way home: you start with one destination but end in another.

Here's how it works:

• The first thing to do is calculate the inner function \( f(x) \), which tells us our initial stopping point.
• Next, using this result as input, you apply the outer function \( g(x) \).
• The result is \( g(f(x)) \), the composite function.

In our problem, the composition involves the functions \( f(x) = x^2 \) and \( g(x) = x - 3 \).
You first calculate \( f(x) = x^2 \), and then use this in \( g(x) = x - 3 \). This layering creates the new function \( g(f(x)) \) = \( x^2 - 3 \). Always remember: the order matters! You can't switch \( f \) and \( g \) without changing the result.

Substitution Method

The Substitution Method is a magical tool for finding composite functions. Imagine it as replacing parts in an equation with something new that fits perfectly. Here, you're substituting the output of one function directly into another, just like changing a single part without affecting the whole system.

Let’s break this down using our example:

• You start with \( f(x) = x^2 \).
• Substitute this into \( g(x) = x - 3 \).
• Now the expression becomes \( g(f(x)) = g(x^2) \).
• This leads to \( (x^2) - 3 \), the simplified form of the composite \( g(f(x)) \).

Using substitution, you seamlessly integrate one function's result into another, ensuring a smooth transition and accurate calculation.
This method is crucial when dealing with variables that are hard to break down into simpler parts, just like assembling a puzzle piece by piece.

Polynomial Functions

Polynomial Functions are like a trusty old bicycle; simple yet incredibly useful. They are mathematical expressions involving sums of powers of \( x \) with coefficients. These functions are easy to handle and ideal for mathematical operations, including composite functions.

In our example, the polynomial function \( f(x) = x^2 \) comes into play. Functions of this kind have specific characteristics:

• They are usually expressed in terms of \( x^n \) where \( n \) is a non-negative integer.
• Each term involves coefficients, which are real numbers (like the \( 1 \) in \( 1x^2 \) for \( f(x) \)).
• They can be simple, like \( x \) or more complex, like \( x^3 + 2x^2 - 5 \).

When you substitute into a polynomial, each term remains intact, making polynomial calculations with composite functions straightforward.
This is why polynomial functions are frequently the backbone of many algebraic practices and give you a firm understanding of basic function behavior.