Problem 2 - Brief Applied Calculus

Question

Find functions \(f\) and \(g\) such that the given function is the composition \(f(g(x))\). $$ \left(5 x^{2}-x+2\right)^{4} $$

Step-by-Step Solution

Answer

Functions are \( f(u) = u^4 \) and \( g(x) = 5x^2 - x + 2 \).

Step 1: Identify the Outer Function

First, look at the structure of the given function: \( \left(5x^2 - x + 2\right)^4 \). Notice that there is an exponent "4" applied to the entire expression. This suggests that the outer function, \( f(x) \), is something raised to the power of four. Therefore, we can define the outer function as \( f(u) = u^4 \), where \( u \) represents the inner function.

Step 2: Identify the Inner Function

Next, examine what is inside the parentheses in the expression \( \left(5x^2 - x + 2\right)^4 \). This expression \( 5x^2 - x + 2 \) forms the inner part of the composite function. Therefore, we can define the inner function \( g(x) \) as \( g(x) = 5x^2 - x + 2 \).

Step 3: Construct the Composite Function

Using the functions identified in the previous steps, construct the composite function \( f(g(x)) \). Substitute \( g(x) = 5x^2 - x + 2 \) into the outer function \( f(u) = u^4 \). Therefore, the composite function \( f(g(x)) \) is equivalent to \( (5x^2 - x + 2)^4 \), which matches the original function given.

Key Concepts

Outer Function

When you're dealing with function composition, the outer function is applied last, but it determines the overarching structure of the expression. In the example function \((5x^2 - x + 2)^4\), you start by noticing how the entire expression is raised to the fourth power. This exponentiation is what defines your outer function. The outer function basically acts like a framework. It's the structure within which the inner function operates.
For our specific problem, the outer function is defined as \(f(u) = u^4\). Here, the variable \(u\) acts as a place-holder for whatever the inner function might be. Think of it as a shell or a casing that wraps around the actual content—like a gift inside a box. The outer function, therefore, is critical because it transforms the input it receives based on its own formula, which in this case is lifting the input to the fourth power.

• The outer function takes its argument—inside the composite—and raises it to a power.
• Helps define the final shape and behavior of the composite function.
• Acts as a finale for the expression evaluation within a composition.

Inner Function

The inner function is the part of a composition that gets evaluated first. It's usually found within the parentheses of a composite function. In our exercise, this is the term inside \((5x^2 - x + 2)^4\). By evaluating this piece first, we lay the groundwork for what the outer function will act upon.
In simpler terms, the inner function does the initial processing. For our example, it is represented as \(g(x) = 5x^2 - x + 2\). This function takes an input \(x\) and performs the operations: multiplying by 5, squaring, subtracting \(x\), and then adding 2.
The beauty of the inner function lies in its ability to transform input values into something that the outer function can further manipulate.

• Evaluated first when computing a composite function.
• Acts as a foundation for the subsequent application of the outer function.
• May take a variety of forms such as polynomial, trigonometric, etc.

Composite Function

The composite function is the practical application of combining two or more functions in a nested manner. In our exercise, you combine the insights from both the inner and outer functions to create a unified expression. This results in a new single function.
For the example \((5x^2 - x + 2)^4\), the composite function is denoted as \(abla f(g(x))\). It starts with computing \(g(x) = 5x^2 - x + 2\) and then feeding this result as an input to \(f(u) = u^4\), ultimately combining to give the function \(f(g(x)) = (5x^2 - x + 2)^4\).
Composite functions provide a powerful way of working with multiple operations at once. This can be used to create intricate and complex expressions with relative ease.

• Results from plugging the output of one function directly into another.
• Allows you to build complex expressions from simpler ones.
• Maintains the order: evaluate the inner function first, followed by the outer function.