Problem 93 - Algebra and Trigonometry

Question

Solving an Equation for an Unknown Function In Exercise 69 of Section 3.6 you were asked to solve equations in which the unknowns were functions. Now that we know about inverses and the identity function (see Exercise 92 ) we can use algebra to solve such equations. For instance, to solve \(f \circ g=h\) for the unknown function \(f,\) we perform the following steps: $$ \begin{aligned} f \circ g &=h \\ f \circ g \circ g^{-1} &=h \circ g^{-1} \\\ f \circ I &=h \circ g^{-1} \\ f &=h \circ g^{-1} \end{aligned} $$ So the solution is \(f=h \circ g^{-1} .\) Use this technique to solve the equation \(f \circ g=h\) for the indicated unknown function. (a) Solve for \(f,\) where \(g(x)=2 x+1\) and $$ h(x)=4 x^{2}+4 x+7 $$ (b) Solve for \(g,\) where \(f(x)=3 x+5\) and $$ h(x)=3 x^{2}+3 x+2 $$

Step-by-Step Solution

Answer

(a) The function \( f = x^2 + 6 \). (b) The function \( g = x^2 + x - 1 \).

Step 1: Understand the Composition

The problem states that you have a composition of functions, specifically \( f \circ g = h \), meaning \( f(g(x)) = h(x) \). The goal is to solve for one of the functions given the others.

Step 2a: Find the Inverse of g (Part a)

Given \( g(x) = 2x + 1 \), find its inverse \( g^{-1}(x) \). To do this, set \( y = 2x + 1 \) and solve for \( x \):\[ y = 2x + 1 \ y - 1 = 2x \ x = \frac{y - 1}{2} \] So, \( g^{-1}(x) = \frac{x - 1}{2} \).

Step 3a: Substitute to Solve for f (Part a)

Using the equation \( f \circ g = h \), we have \( f = h \circ g^{-1} \). Substitute \( g^{-1}(x) = \frac{x - 1}{2} \) into \( h \):\[ h(g^{-1}(x)) = 4\left(\frac{x - 1}{2}\right)^2 + 4\left(\frac{x - 1}{2}\right) + 7 \ = 4 \cdot \frac{(x - 1)^2}{4} + 2(x - 1) + 7 \] Simplify:\[ = (x - 1)^2 + 2(x - 1) + 7 \] Further simplification gives:\[ = x^2 - 2x + 1 + 2x - 2 + 7 \] \[ = x^2 + 6 \] So, \( f(x) = x^2 + 6 \).

Step 2b: Find the Inverse of f (Part b)

Given \( f(x) = 3x + 5 \), find its inverse \( f^{-1}(x) \). Set \( y = 3x + 5 \) and solve for \( x \):\[ y - 5 = 3x \ x = \frac{y - 5}{3} \]Thus, \( f^{-1}(x) = \frac{x - 5}{3} \).

Step 3b: Substitute to Solve for g (Part b)

Using the equation \( f \circ g = h \), we need to express \( g = f^{-1} \circ h \). Substitute \( f^{-1}(x) = \frac{x - 5}{3} \) into \( h \):\[ f^{-1}(h(x)) = \frac{3x^2 + 3x + 2 - 5}{3} \] Simplify:\[ = \frac{3x^2 + 3x - 3}{3} \] \[ = x^2 + x - 1 \] So, \( g(x) = x^2 + x - 1 \).

Key Concepts

Function Composition

When dealing with inverse functions and solving for an unknown function, it's crucial to understand function composition. Composition of functions, denoted as \(f \circ g\), implies applying one function to the results of another. In this context, \(f \circ g\) equates to \(f(g(x))\). Here, your goal is to determine either \(f\) or \(g\), given the other functions. The concept is similar to a pipeline, where \(g(x)\) processes an input, and then the output is processed by \(f(x)\). This is an essential mechanism when you need to decompose or reconstruct complex expressions into simpler, manageable parts.
Function composition also helps understand the layering of operations within an equation, where you perform a sequence of functions. This is highly beneficial when identifying missing components or expressing one function in terms of another. Remember, function composition isn't just combining operations; it encapsulates them into a flow that maintains the integrity of mathematical transformations.

Solving Equations

Solving equations with inverse functions introduces a dynamic approach to finding unknowns. It builds on finding the inverse of a function to reverse its effect. This means identifying a function that 'undoes' the original function's processing. When solving \(f \circ g = h\), acknowledging the role of each component is essential:

• Identify each function and its role in the composition.
• Find the inverse of the known functions, when necessary, to solve for the unknowns.
• Use these inverses to isolate the unknown function.

This involves substituting back and simplifying equations. For instance, if you know \(g(x)\), you find \(g^{-1}(x)\) to substitute into \(h(x)\), giving you the form of \(f(x)\). In particular, the key step is setting \(f = h \circ g^{-1}\) and simplifying to uncover \(f(x)\).'"Ensure clarity with each substitution and follow-through logically to the solution.

Algebraic Manipulation

In problems with inverse functions and compositions, algebraic manipulation facilitates the transformation and simplification of expressions. This process involves techniques to rearrange equations, isolate variables, and simplify terms. Here's why it's critical:

• You often switch between different forms to identify or transform functions, like moving from \(f \circ g\) to \(f = h \circ g^{-1}\).
• To find inverses, you switch dependent and independent variables and resolve for the original input.
• When substituting, carefully navigate through each operation: squaring, expanding, or simplifying coefficients. As shown in simplifying \( \left( \frac{x - 1}{2} \right)^2 \).

In essence, mastery over algebraic manipulation ensures you keep track of the interplay between variables and functions. It enables seamless transitions from complex to simplified forms, making it easier to spot solutions and verify results. This skill, along with understanding inverse functions, equips you to tackle diverse problems in mathematics effectively.