Problem 96 - Precalculus : Building Concepts and Connections

Question

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(3 x-7)^{10}+5(3 x-7)^{2}$$

Step-by-Step Solution

Answer

So, the two functions \( f \) and \( g \) such that \( h(x) = f(g(x)) \) are \( f(x) = x^{10} + 5x^{2} \) and \( g(x) = 3x - 7 \).

Step 1: Express h(x) as a composite function

Firstly, notice that inside the term under both powers (10 and 2) in h(x), there is a function \( g(x) = 3x - 7 \). This suggests that the given function is a composition of a function of \( g(x) \) and the outside function.

Step 2: Identify the outer function

After identifying the inner function as \( g(x) = 3x - 7 \), next is to identify the outer function \( f \). The outer function acts on the result of \( g(x) \). In this case, it takes \( g(x) \), elevates it to power 10, adds to it 5 times the square of \( g(x) \). So the outer function can be represented as \( f(x) = x^{10} + 5x^{2} \).

Step 3: Show the composite function

Now that both functions \( f \) and \( g \) have been identified, \( h(x) \) can be rewritten as the composite of \( f \) and \( g \). That gives \( h(x) = f(g(x)) = (3x - 7)^{10} + 5(3x - 7)^{2} \)

Key Concepts

Function Composition

Talking about function composition is like discussing a recipe with multiple steps. Each step modifies the ingredients, resulting in a final product that's affected by every process it underwent. In mathematics, function composition combines two or more functions into a single function, where the output of one function becomes the input of the next.

Take, for example, the exercise given which presents a composite function denoted as \( h(x) = (f \circ g)(x) = f(g(x)) \). Here, the function \( g(x) \) is applied first, and then function \( f \) is applied to the result of \( g(x) \) to create a new function \( h(x) \). This process turns two individual functions into a teamwork effort, resulting in a final expression that reflects the combined actions of both \( f \) and \( g \).

By identifying two functions, \( f \) and \( g \) that make up \( h(x) \) as in the solution steps, students witness the unfolding of the composite function. This understanding clarifies how complex functions can actually be a series of simpler functions working together.

Inner and Outer Functions

When we compose functions, we refer to the function that gets applied first as the 'inner function' and the function that gets applied second as the 'outer function'. This terminology helps us keep track of the order in which operations occur during the function composition process.

From the exercise at hand, \( g(x) = 3x - 7 \) is clearly the inner function. It's the foundation on which further transformation is built. Think of it as the core ingredient in our recipe that's prepped before anything else is added. The outer function, in this case presented as \( f(x) = x^{10} + 5x^{2} \), then processes the outcome of the inner function to generate the final result.

A good way to visualize this is by imagining the inner function as nestled within the outer function. The smooth operation of combining the two functions is equivalent to ensuring the core ingredient (inner function) blends perfectly within the overarching recipe (outer function) to produce a delicious dish (the composite function).

Exponentiation

Exponentiation is a mathematical operation that involves raising a number, known as the base, to the power of an exponent. It signifies repeated multiplication, much like how layering ingredients can intensify flavors in cooking. When a base is raised to an exponent, it's multiplied by itself as many times as specified by the exponent value.

In the context of our exercise, exponentiation is seen in the terms \( (3x - 7)^{10} \) and \( (3x - 7)^{2} \). The involvement of exponentiation indicates a more elaborate transformation than just a simple addition or multiplication within the function composition. This is where the power of exponentiation becomes evident—by drastically changing the behavior of the equation with each increase in the exponent value.

Understanding exponentiation within function composition is like recognizing how spices can change the dynamics of a dish. Just as a pinch or a heap of spice can make a significant difference, so does changing the exponent in a function.