Question

Identify the inside function, \(u=g(x)\), and the outside function, \(y=f(u)\). $$ y=f(g(x)) \quad u=g(x) \quad y=f(u) $$ $$ y=(6 x-5)^{4} $$

Short answer

The inside function, \(g(x) = 6x - 5\), and the outside function, \(f(u) = u^{4}\).

Step-by-step solution

Identify the composite function

Recognize the given function as a composite function. The function given is \(y=(6x-5)^{4}\). Here, you can see that \(y\) is a function of another function (i.e., \(6x-5\) which is raised to the 4th power). Hence, we can consider the given function \(y\) as a composition of two functions.

Identify the inside function (\(g(x)\)

The inside function, \(g(x)\), is the part of the function that is acted upon first. In this case, it is the \(6x-5\) part of the given function. As it is being raised to a power (4), this means we first perform the operation \(6x-5\). So, our \(g(x) = 6x - 5\). It gets substituted into the other function. This is our inside function or \(g(x)\).

Identify the outside function (\(f(u)\)

The outside function, \(f(u)\), is the function that is acting on the inside function (\(g(x)\)). In this case, it happens to be the function that is raising a quantity to the 4th power. Thus, the outside function or \(f(u)\) is \(u^4\), which will act on the result of \(g(x)\). Or you can understand it as \(f(u)=(u)^{4}\).

Key concepts

Inside Function

When we talk about composite functions, the term 'inside function' represents the part of the composition that is computed first. It's pretty much the foundation for the rest of the function to build upon. In the example provided, the inside function is identified by the equation \( g(x) = 6x - 5 \).

Think of it as the 'core' of an apple. Before you can appreciate the taste of the whole apple, you need to get to its core. Similarly, in function composition, before applying further transformations, you calculate the inside function. Once you've evaluated this part, you then input its result into the 'outside function' for further processing.

Outside Function

The 'outside function' can be seen as the next layer that wraps around the inside function—it's what you apply to the result of the inside function. Continuing with our example where \( g(x) \) was our inside function, the outside function is expressed as \( f(u) = u^4 \).

In visual terms, if the inside function is the core of an apple, the outside function is the rest of the apple around the core. This outside part gets impacted by the state of the core—whatever we find when calculating the inside function, we then elevate to the power of four as indicated by the outside function.

Function Composition

Function composition is the fascinating algebraic operation where you combine two functions to create a new one. It's where the output of one function becomes the input for another. Essentially, it's like a relay race—once one runner (the inside function) finishes, it hands the baton to the next runner (the outside function).

In the equation \( y = f(g(x)) \), you're essentially feeding the output of \( g(x) \) directly into the function \( f(u) \), producing a third function \( y \) that shows how \( x \) and \( y \) are related through this two-step process. The beauty of function composition lies in its ability to create complex relationships between variables through a series of simpler functions.

Algebraic Operations

Algebraic operations are the bread and butter of understanding and manipulating functions. They are the set of rules and procedures that allow us to perform calculations with unknown values. In the context of our earlier example, raising the inside function to the fourth power is an algebraic operation. It shows us how we can transform a simple linear function \( g(x) = 6x - 5 \) into something more complex through exponentiation.