Problem 4 - College Algebra and Calculus: An Applied Approach

Question

Identify the inside function, \(u=g(x)\), and the outside function, \(y=f(u)\). $$ y=f(g(x)) \quad u=g(x) \quad y=f(u) $$ $$ y=\left(x^{2}+1\right)^{4 / 3} $$

Step-by-Step Solution

Answer

The inside function \(u=g(x)\) is \(x^{2}+1\), and the outside function \(y=f(u)\) is \(u^{4 / 3}\).

Step 1: Identify the Composite Function

In the expression given, \(y=(x^{2}+1)^{4 / 3}\), we can see that it's a composite function. A composite function is a function composed of two or more functions. It's represented in the form \(f(g(x))\), where \(g(x)\) is the inside function and \(f(u)\) is the outside function. To ascertain these, let's first identify the inner function (inside function).

Step 2: Identify the Inside Function

We want to identify the function that is operated on, or the inside function. Inspecting \(y=(x^{2}+1)^{4 / 3}\), we can see that \(x^{2}+1\) is in the inner part of the given function, and it's being acted on with the power of \(4/3\). Hence, the inside function \(u=g(x)\) is \(g(x)=x^{2}+1\).

Step 3: Identify the Outside Function

Now let's identify the function that is performing the operation, or the outside function. This is the operation applied to \(u=g(x)\). In our function, after finding our \(g(x)\), the remaining operation is taking the power of \(4/3\). Hence, the outside function is \(y=f(u)=u^{4 / 3}\).

Key Concepts

Inside Function

When dealing with composite functions, the "inside function" is the function encapsulated within another function. It's the one that receives the initial input or variable. In our example, we have the expression \( y = (x^{2} + 1)^{4/3} \). Here, the inside function is \( g(x) = x^{2} + 1 \).

The inside function acts as a foundation for building the result of the composite function. To better understand, imagine the process as baking a cake where \( g(x) \) is like mixing ingredients. These ingredients, or the output of the inside function, \( g(x) \), will later undergo another process.

Identifying the inside function is crucial because it tells us what happens first in the sequence of operations. In mathematical notation, if you have \( f(g(x)) \), always look for the expression inside the innermost parentheses first. This function often translates the original input \( x \) into a form that will then be manipulated further.

Outside Function

The outside function, in composite functions, is responsible for performing the final transformations on the output of the inside function. It modifies the result of the inside function to generate the final result of the composite function. In our case, we are looking at the expression \( y = (x^{2} + 1)^{4/3} \).

After finding the inside function \( g(x) = x^{2} + 1 \), the next step is identifying what operation is made on \( g(x) \). For this example, the outside function is \( f(u) = u^{4/3} \), where \( u \) represents \( g(x) \).

Think of the outside function as the final stage in a manufacturing line. It's about applying the finishing touches, like baking the batter into a cake. Once you comprehend what the inside function returns, the outside function skillfully applies its transformation. In notation, look beyond the parentheses to perceive what additional operations are applied. This helps unravel the complete expression, producing the eventual output.

Function Composition

Function composition is like a mathematical assembly line, where you combine multiple functions to build a new function. The end result is a streamlined process where the output of one function becomes the input to the next. For example, if you have \( y = f(g(x)) \), it describes how \( g(x) \) wraps the original input \( x \) into an intermediate output, and \( f(u) \) completes the process.

In simple terms, think of each function as a step in completing a larger task. Just like when cooking, you follow each step in a recipe. For our given function \( y = (x^{2} + 1)^{4/3} \), \( g(x) = x^{2} + 1 \) is the first step that creates an intermediate product. Then, \( f(u) = u^{4/3} \) acts as the second step, finishing the entire process.

Understanding function composition allows you to see how complex tasks can be deconstructed into simpler, more manageable components. This concept is widely applicable in both math and real-world problems, where breaking things down into smaller parts makes them easier to solve and understand. Once you practice setting up and dissecting composite functions, you'll be able to tackle even the most intricate equations with confidence.