Problem 12 - Calculus Single Variable

Question

Write the function \(h\) as the composition \(h=g \circ f\) of two functions. (There is more than one correct way to do this.) \(h(x)=2 x+7\)

Step-by-Step Solution

Answer

One composition could be \(h(x) = (2(x)) + 7 = g(f(x))\) with \(f(x) = x\) and \(g(x) = 2x + 7\).

Step 1: Understanding the Composition of Functions

To express a function as a composition of two functions, we find two simpler functions, \(f(x)\) and \(g(x)\), such that applying \(f\) first to \(x\), and then applying \(g\) to the result, gives \(h(x)\)

Step 2: Analyzing the Function Form

Given \(h(x) = 2x + 7\), identify components that could be results of simpler functions combined. One possibility is viewing \(h(x)\) as involving initial transformation of \(x\) (such as changing by a factor or shifting) followed by another operation.

Step 3: Choosing the Inner Function \(f(x)\)

Decide what the function \(f(x)\) does first. In this case, let \(f(x) = x + c\). A simple choice is to make \(c = \frac{7}{2}\), leading to \(f(x) = x\) since we will handle the constant in \(g(x)\).

Step 4: Defining the Outer Function \(g(x)\)

Since \(f(x) = x\), we simply rewrite 2x + 7 as an outer function that affects the result of \(f(x)\). Set \(g(x) = 2x + 7\), meaning that \(g(x)\) receives \(f(x)\) as input, fulfilling \(g(f(x)) = 2f(x) + 7\).

Step 5: Composition Verification

To verify, apply \(f(x)\) first to \(x\) giving \(f(x) = x\), and then apply \(g(x) = 2x + 7\) to obtain \(g(f(x)) = 2f(x) + 7 = 2x + 7\). This matches \(h(x)\), completing the composition.

Key Concepts

composition of functions

Composition of functions is a core aspect of understanding how complex functions can be broken down into simpler ones. When we talk about composing functions, it simply means plugging one function into another. Consider two functions, \(f(x)\) and \(g(x)\). If we compose \(f\) and \(g\), we get a new function \(h(x) = g(f(x))\).
To visualize:

• First, you apply \(f\) to \(x\).
• Then, you take the result from \(f(x)\) and apply \(g\) to it.

The result is that a single value of \(x\) is transformed twice, thus forming a more complex function \(h\).
This concept is very useful for simplifying problems where applying one function to the result of another function clarifies the structure. In mathematical notation, \(g \circ f\) denotes this lawful process of composition.

algebraic functions

Algebraic functions are expressions constructed from polynomial equations, using basic algebraic operations like addition, subtraction, multiplication, division, and root extraction. For the given problem, we are considering the function \(h(x) = 2x + 7\). This is a linear algebraic function. It multiplies \(x\) by a constant (in this case, 2), and adds another constant (7).
Linear functions such as this one are foundational because:

• They are the simplest form of algebraic functions.
• They provide a straight-line graph, making them predictable and easy to analyze.
• They serve as building blocks for more complex functions, acting as components in function composition.

In our problem, rewriting \(h(x)\) as the composition of two functions \(f(x)\) and \(g(x)\) involves simplifying the performing of operations on \(x\) and helps to understand how different algebraic manipulations combine to give a function its form.

function transformation

Function transformation involves changing a function's graph by manipulating its algebraic form. Typically, transformations include shifts, stretches, compressions, and reflections. In our function \(h(x) = 2x + 7\), consider how the components of this function represent transformations:

• \(2x\) signifies a vertical stretch by factor 2: each \(y\)-value of a base function \(x\) is multiplied by 2.
• Adding 7 results in a vertical shift: the entire graph moves up by 7 units.

These transformations affect the graph and the function's behavior. When decomposing \(h(x) = 2x + 7\) into compositions like \(g(f(x))\), we break down these transformations into manageable pieces. Such practices help to grasp each step's impact on the resulting function, making transformations easier to visualize and understand. This clarifies how altering algebraic expressions translates into geometrical manipulations of graphs.