Problem 69 - Algebra and Trigonometry

Question

Solving an Equation for an Unknown Function Suppose that $$ \begin{aligned} g(x) &=2 x+1 \\ h(x) &=4 x^{2}+4 x+7 \end{aligned} $$ Find a function \(f\) such that \(f \circ g=h\) . (Think about what operations you would have to perform on the formula for \(g\) to end up with the formula for \(h\) .) Now suppose that $$ \begin{array}{l}{f(x)=3 x+5} \\ {h(x)=3 x^{2}+3 x+2}\end{array} $$ Use the same sort of reasoning to find a function \(g\) such that \(f \circ g=h\)

Step-by-Step Solution

Answer

Function \(f(x) = x^2 + 2x + 5.5\) with \(g(x) = x\) forms the correct compositions.

Step 1: Understand the Composition

The composition of two functions, such as \(f \circ g\), means that for every input \(x\), the output of \(g(x)\) becomes the input for \(f(x)\). In our situation, we need to find \(f\) such that \(f(g(x)) = h(x)\).

Step 2: Composition for g(x) = 2x + 1

Since \(f(g(x)) = h(x)\) and \(h(x) = 4x^2 + 4x + 7\), substitute \(g(x)\) in \(h(x)\) to find \(f(x)\): Set \(y = g(x) = 2x+1\). Then replace \(x\) in \(h(x)\) with \(y\): \[\begin{align*} h(y) &= 4\left(\frac{y-1}{2}\right)^2 + 4\left(\frac{y-1}{2}\right) + 7.\end{align*}\] Simplifying gives \(f(y) = y^2 + 2y + 5.5\). So, \(f(x) = f(g(x)) = 4x^2 + 4x + 7\).

Step 3: Simplify h(y) with Substitution

Substitute \(g(x)\) as \(2x+1\) into the expression for \(h(x)\) to verify the composition: \(f(2x+1) = 4x^2 + 4x + 7\) confirms \(f(x) = (\frac{x-1}{2})^2 + 2(\frac{x-1}{2}) + 5.5\).

Step 4: Use Composition for the Second Problem

Here, \(f(x) = 3x+5\) and \(h(x) = 3x^2 + 3x + 2\). We want \(f(g(x)) = h(x)\). First, consider \(y = g(x)\) and set \(f(y) = 3y + 5\). If \(h(x) = 3x^2 + 3x + 2\), try rewriting this as \(f(g(x)) = 3(g(x))^2 + 3g(x) + 2\).

Step 5: Identify g(x)

To make \(f(g(x)) = h(x)\), you need to choose \(g(x)\) so that when substituted into \(f(x) = 3x+5\), it gives the same expression as \(h(x)\), i.e., \(3y^2 + 3y + 2\). Therefore, solve: \((3y + 5)^2 = 3x^2 + 3x + 2\). By rearranging and factoring, determine that \(g(x) = x\).

Key Concepts

Solving Equations

Solving equations often involves finding a particular value or set of values that satisfy the given relationship or equation. In the context of functions, this means finding a function that will transform one input to match another. When dealing with the composition of functions, such as solving for an unknown function when given a composition like \( f \circ g = h \), several steps are essential.

• First, you understand what each function in the composition is doing individually.
• Then, identify the unknown part and express it in terms of the known functions.
• To solve the equation, you will need to manipulate it so the unknown function is isolated or clearly described.

Each step in solving involves different layers of processing, from substitution to simplification, ensuring that the composition holds true according to the problem's conditions. By solving these types of equations, you gain insight into how separate functions work together to transform inputs and outputs. This skill is essential for higher-level mathematics and useful in various applications, including physics and engineering.

Unknown Functions

Understanding and finding unknown functions is crucial in mathematics, especially when you need to deduce how an output function comes from an input one. When presented with a composition like \( f \circ g = h \), you must carefully consider each function's role. For instance, if you know \( g(x) \) and \( h(x) \), finding \( f(x) \) means discerning what transformations \( g(x) \) undergoes to become \( h(x) \).

• This requires you to substitute and replace variables correctly, turning the operation inside out, much like solving for an unknown in an algebraic equation.
• You start by expressing each component with known values, then logically deducing how the unknown function bridges the gap between \( g(x) \) and \( h(x) \).
• The ultimate goal is to write the unknown function in a simplified and resolved form.

To find an unknown function within a composition, always think about what changes each step in the transformation applies and ensure each corresponds correctly to produce the desired results.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and solving equations to find unknowns, which often requires various techniques such as substitution, expansion, and simplification. In function composition problems, algebraic manipulation is an invaluable tool.

• Start by identifying what mathematical operations are needed to isolate and express the unknown functions.
• Use substitution to replace known expressions with these unknowns, allowing you to work within familiar structures.
• Keep simplifying the expressions until the unknowns are isolated or expressed by the known parts.

In the example of finding a function \( f \) for \( f \circ g = h \), algebraic manipulation helps you substitute \( g(x) \) into \( h(x) \), simplify the expression, and thus identify the form of \( f(x) \). Through practice, you will recognize patterns, understand transformations, and develop confidence in using algebraic manipulation to solve complex equations. This strengthens your overall mathematical problem-solving skills, making it easier to tackle various mathematical challenges.