Problem 76 - Precalculus Essentials

Question

Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=(2 x-5)^{3}$$

Step-by-Step Solution

Answer

The function \(h(x)=(2x-5)^3\) can be expressed as a composition of two functions: \(f(x) = x^3\) and \(g(x) = 2x - 5\). So, \(h(x)=(f \circ g)(x)\).

Step 1: Identify the Inner Function g(x)

Look at the function \(h(x) = (2x-5)^3\). The first function \(g(x)\), is the inner function which takes \(x\) as an input and outputs \(2x - 5\). So we have \(g(x) = 2x - 5\).

Step 2: Identify the Outer Function f(x)

The second function \(f(x)\) takes the result of \(g(x)\) and operate on it. In this case, \(f(x)\) is the function that cubes its input. Therefore, \(f(x) = x^3\).

Step 3: Validate the Decomposition

Verify that \(h(x)=(f \circ g)(x)\) by composing \(f\) and \(g\). When we perform the function composition, we get \(f(g(x))=f(2x-5)=(2x-5)^3\), which confirms that our chosen functions \(f\) and \(g\) are correct.

Key Concepts

Inner and Outer Functions

Understanding the concept of inner and outer functions is crucial when dealing with function composition, a fundamental operation in mathematics. In the context of our exercise, where we have the function h(x) = (2x-5)^3, this concept becomes tangible.

Imagine you're getting ready for a formal event. You put on your clothing in layers: the first layer (inner) and then your coat (outer). Functions work similarly in composition. The first function that operates directly on the input, x, is your inner function, designated as g(x). Think of it like putting on your inner layer of clothes. In our example, g(x) = 2x - 5.

Next, comes the outer function, f(x), which takes the output of your inner function (just like you would put on a coat over your clothes) and does something further with it. In the case at hand, f(x) raises its input to the third power, hence f(x) = x^3. The composition (f \(circ\) g)(x) can be visualized as applying one function after the other, in the exact order of dressing up. This helps simplify complex expressions or understand how combined operations work on a given input.

To aid comprehension, it's beneficial to practice by identifying the inner and outer functions of different compositions and to recognize how the output from the inner function seamlessly becomes the input for the outer function.

Function Decomposition

The concept of function decomposition is akin to breaking down a complex task into more manageable steps—much like following a recipe step-by-step to make a complicated dish. Taking the function h(x) = (2x-5)^3, function decomposition involves expressing h(x) as a combination of simpler functions, f(x) and g(x), that when composed, recreate h(x).

In our exercise, we decomposed h(x) into g(x) = 2x - 5, handling the linear transformation of x, and f(x) = x^3, executing the cubic transformation of the output from g(x). This piecemeal approach offers a clearer view of a complex function's workings by isolating its parts, facilitating easier computation and analysis.

To strengthen this understanding, students can practice decomposing various composite functions. This way, they can better grasp how to dismantle complex functions into more fundamental ones, which is a critical skill in calculus and higher-level math.

Function Operations

Diving into function operations, we explore the different ways functions can be combined to create new functions. Specifically, through operations like composition, addition, multiplication, and division, we can build complex functions from simpler ones. Consider our initial example where h(x) = (2x-5)^3 was expressed through function composition.

Function composition, (f \(circ\) g)(x), is a specific operation where you apply one function to the result of another—like putting on socks before shoes. Additional operations like adding (f + g)(x) = f(x) + g(x), or multiplying functions (fg)(x) = f(x)g(x), all lead to new functions with their own unique properties.

Getting to grips with these operations is not just about going through the motions; it’s about seeing the interplay between functions and understanding how complex relationships between variables can be structured. To enrich their learning, students should practice these operations with a variety of functions to see firsthand the vast possibilities functions hold when combined in different ways.