Problem 29 - Calculus with Concepts in Calculus

Question

Write \(h\) as the composite \(g \circ f\) of two functions \(f\) and \(g\) (neither of which is equal to \(h\) ). $$ h(x)=\left(3 x^{2}-5 \sqrt{x}\right)^{1 / 3} $$

Step-by-Step Solution

Answer

Define \( f(x) = 3x^2 - 5\sqrt{x} \) and \( g(u) = u^{1/3} \), then \( h(x) = g(f(x)) \).

Step 1: Identify the Outer Function

The given function is \( h(x) = \left(3x^2 - 5\sqrt{x}\right)^{1/3} \). To express this as a composition \( g(f(x)) \), we first identify the outer function. The outer transformation is taking the cube root, \( y^{1/3} \). Thus, define \( g(u) = u^{1/3} \) where \( u = f(x) \).

Step 2: Determine the Inner Function

Next, consider the expression inside the cube root in \( h(x) \), which is \( 3x^2 - 5\sqrt{x} \). This will be our inner function. Define & justify \( f(x) = 3x^2 - 5\sqrt{x} \). Now, \( h(x) = g(f(x)) \) means \( g(f(x)) = \left(3x^2 - 5\sqrt{x}\right)^{1/3} \), consistent with the given \( h(x) \).

Step 3: Verify the Composition

Now let's verify our choices by substituting \( f(x) \) into \( g(u) \). If \( f(x) = 3x^2 - 5\sqrt{x} \), substituting into \( g \) gives \( g(f(x)) = (3x^2 - 5\sqrt{x})^{1/3} \). Since this is the original \( h(x) \), the decomposition is correct.

Key Concepts

Function Composition

Function composition is a powerful way to build new functions by combining existing ones. It's like creating a recipe where one function is the main ingredient of another. If you have two functions, say \( f(x) \) and \( g(x) \), you can create a new function \( h(x) \) by substituting \( f(x) \) into \( g(x) \). This is written as \( g(f(x)) \) or sometimes using the composition symbol as \( g \circ f \). The result is called a composite function.
Composing functions allows you to encapsulate complex operations in a simple expression. It makes solving and analyzing equations much easier, as you can break down complicated functions into simpler parts.

• How it works: Evaluate the inner function first, then apply the outer function to the result.
• Purpose: Simplifies equations and operations by turning a series of steps into one operation.

Inner Function

In the context of function composition, the inner function is the function that is evaluated first, before its result is passed to the outer function. If given a function \( h(x) = g(f(x)) \), here the inner function is \( f(x) \).
For the original exercise, the inner function was identified as \( f(x) = 3x^2 - 5\sqrt{x} \). This expression represents the operations that occur within the cube root in \( h(x) \).
By identifying the inner function, you are recognizing the first step in the overall composition, setting the stage for the larger function operation performed by the outer function.

• Characteristics: It is the first function to be evaluated in a composite function.
• Role: Provides the input for the outer function.

Outer Function

The outer function in a composition is the one that takes the output of the inner function as its input. It acts as the final step in the computation chain of a composite function. In cases where you see \( h(x) = g(f(x)) \), \( g(x) \) is the outer function.
In our example, \( g(u) = u^{1/3} \) was the outer function, representing the cube root transformation. It takes the result of \( f(x) = 3x^2 - 5\sqrt{x} \) and extends it by performing the cube root operation.
Understanding the outer function helps in predicting and understanding the behavior of the composite function, as it provides the final transformation on the output from the inner function.

• Main Functionality: Transforms the result of the inner function.
• Importance: Completing the computation and defining the final output of the composite function.