Problem 30 - College Algebra

Question

Find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\). $$ h(x)=4+\sqrt[3]{x} $$

Step-by-Step Solution

Answer

Let \( f(u) = 4 + u \) and \( g(x) = x^{1/3} \).

Step 1: Identifying the Inner Function

The given function is \( h(x) = 4 + \sqrt[3]{x} \). We identify \( g(x) \) as the inner function. A common feature of \( h(x) \) is the \( \sqrt[3]{x} \) term, so we assign \( g(x) = \sqrt[3]{x} = x^{1/3} \).

Step 2: Defining the Outer Function

The outer function \( f(x) \) will take the form of the simplest representation of \( h(x) \) when \( g(x) \) is inserted. Set \( f(u) = 4 + u \), where \( u = g(x) = \sqrt[3]{x} \).

Step 3: Verification

Verify that \( h(x) = f(g(x)) \). Substitute \( g(x) = \sqrt[3]{x} \) into \( f(u) \) to get \( f(g(x)) = 4 + g(x) = 4 + \sqrt[3]{x} \). This matches the original function \( h(x) \).

Key Concepts

Function Decomposition

Function decomposition involves breaking down a complex function into simpler parts. Imagine it like taking a whole piece of pizza and slicing it into smaller pieces. This process helps us analyze and understand the function better.

When we decompose a function, we look for two components:

• The inner function, which operates first on the input, and
• The outer function, which takes the output from the inner function and processes it further.

By identifying these two parts, we can express a given function as a composition of simpler functions. For example, if we have a function like \( h(x) = 4 + \sqrt[3]{x} \), decomposing it helps us write it in the form \( f(g(x)) \), where \( f(x) \) and \( g(x) \) are these simpler functions.

Outer Function

The outer function is the second part of the function decomposition process. It is the function that wraps around the output of the inner function and applies additional operations.

In our example, we want to simplify \( h(x) = 4 + \sqrt[3]{x} \). After identifying \( g(x) = \sqrt[3]{x} \) as our inner function, the next step is to define the outer function \( f(u) \). We achieve this by noticing how the inner function plugs into the complete function.

Thus, the outer function is \( f(u) = 4 + u \), where \( u \) represents the output from the inner function. Essentially, \( f(u) \) adds 4 to whatever input it receives from \( g(x) \). Hence, in the broader combination, the role of the outer function is to add that constant value to the result of the cubed root.

Inner Function

The inner function is the first layer of the function decomposition and acts directly on the input \( x \) before the result is passed to the outer function.

Take \( h(x) = 4 + \sqrt[3]{x} \) as our example. By inspecting the expression, we spot the cube root part \( \sqrt[3]{x} \). This segment of the function comes first in action and hence is identified as the inner function.

We define it as \( g(x) = \sqrt[3]{x} \) which is equivalent to \( x^{1/3} \). This means, given an input \( x \), the inner function transforms it by finding its cubed root. The result becomes the input for the outer function, highlighting the sequential nature of function decomposition.

Cube Root Function

The cube root function is a specific mathematical operation that is often used in function decomposition.

In our example, the cube root function is signified by \( \sqrt[3]{x} \) in the expression \( h(x) = 4 + \sqrt[3]{x} \). Here, the cube root function plays a crucial role as the inner function, transforming \( x \) into a smaller part of the overall function.

For any input \( x \), the cube root function \( g(x) = \sqrt[3]{x} \) computes \( x^{1/3} \). This operation extracts the cube root, meaning it finds the number that, when raised to the power of three, returns the original \( x \). This transformed value from the cube root function is then used as input for the outer function, facilitating a seamless combination of operations to reconstruct \( h(x) \).