Problem 15 - Calculus Single Variable

Question

In Exercises \(15-18\), find a function \(g\) such that \(h=g \circ f\) \(h(x)=3 x^{2}+6 x+4, f(x)=x+1\)

Step-by-Step Solution

Answer

The function is \( g(x) = 3x^2 + 3 \).

Step 1: Understand the Composition

We need to find a function \( g(x) \) such that when it is composed with \( f(x) = x + 1 \), the result is \( h(x) = 3x^2 + 6x + 4 \). This means \( g(f(x)) = h(x) \).

Step 2: Substitute and Simplify

Substitute \( f(x) = x + 1 \) into \( h(x) \). Therefore, we rewrite \( h(x) = 3(x + 1)^2 + 6(x + 1) + 4 \).

Step 3: Expand the Expression

First, expand \( (x + 1)^2 \) which gives \( x^2 + 2x + 1 \). Insert this into the equation: \( 3(x^2 + 2x + 1) + 6(x + 1) + 4 \).

Step 4: Distribute and Combine Like Terms

Now distribute the 3: \( 3x^2 + 6x + 3 \). Also distribute the 6: \( 6x + 6 \). Add these to get: \( 3x^2 + 6x + 3 + 6x + 6 + 4 \).

Step 5: Simplify the Polynomial

Combine like terms: \( 3x^2 + 12x + 13 \). Thus we have: \( 3x^2 + 12x + 13 = h(x) \).

Step 6: Determine \( g(x) \)

Since \( g(f(x)) = 3((x+1))^2 + 6(x+1) + 4 \) simplifies to \( g(x) = 3x^2 + 12x + 13 \), we have \( g(x) = 3x^2 + 3 \).

Key Concepts

Polynomials

A polynomial is an expression involving variables and coefficients, composed of terms whose exponents are whole numbers. In the context of our exercise, the function \( h(x) = 3x^2 + 6x + 4 \) is a quadratic polynomial. The terms of this polynomial include:

• \( 3x^2 \), which is the quadratic term.
• \( 6x \), the linear term.
• \( 4 \), the constant term.

Polynomials like this are vital in algebra due to their straightforward nature. They can model various real-world phenomena, and because of this, learning how to manipulate and understand them is crucial.When dealing with polynomials, it's helpful to remember that:

• Each term is separated by a plus (+) or minus (-) sign.
• The highest exponent defines the degree of the polynomial, in this case, 2 (making it quadratic).

Understanding polynomials allows you to perform various algebraic tasks, such as simplifying expressions, finding function compositions, and more.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to reach a desired outcome. In the given exercise, we examined the function composition to rewrite \( h(x) \) by substituting \( f(x) = x + 1 \).We performed several critical algebraic steps:

• Substitution: Replacing \( x \) in \( h(x) \) with \( f(x) = x + 1 \) to find \( g(x) \).
• Expanding: Finding \((x + 1)^2 \), which expands to \(x^2 + 2x + 1\).
• Distribution: Multiplying the expanded terms by the coefficients, such as \(3(x^2 + 2x + 1)\).
• Simplification: Combining like terms to eventually derive \( g(x) = 3x^2 + 12x + 13 \).

Algebraic manipulation is crucial in solving equations, simplifying expressions, and ensuring your solutions remain correct. By using these techniques, you'll also develop a deeper understanding of algebraic relationships and how different mathematical elements interact.

Mathematical Functions

A function is a relation between a set of inputs and a set of possible outputs, where each input is related to exactly one output. The problem in our example demonstrates the concept of function composition, where two functions are combined to produce a new function.Here’s how it works in our exercise:

• Function \( f(x) \): This is a simple linear function given by \( x + 1 \).
• Function \( g(x) \): This is what we are determining to ensure that the composition \( g(f(x)) = h(x) \).
• Function \( h(x) \): Our target polynomial \( 3x^2 + 6x + 4 \).
• Function Composition: We input \( f(x) \) into \( g(x) \) to result in \( h(x) \).

Functions and their compositions allow us to model complex changes and transitions in data or variables. Understanding how to compose functions helps to solve problems efficiently, especially in complex systems where multiple changes occur in succession.