Problem 29 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find (a) \(f \circ g\) and (b) \(g \circ f\). $$f(x)=x^{3}-1, g(x)=\sqrt[3]{x+1}$$

Step-by-Step Solution

Answer

The composite functions \(f \circ g\) and \(g \circ f\) are both \(x\).

Step 1: Function Composition \(f \circ g\)

The composition of the functions \(f\) and \(g\) is denoted as \(f \circ g\). It is the function obtained by performing \(f\) after \(g\) on an input \(x\). In other words, it means substitute \(g(x)\) into \(f(x)\). The formula for function composition is: \(f(g(x))\). Here, substitute \(g(x)=\sqrt[3]{x+1}\) into \(f(x)\), thus \(f(g(x))=(\sqrt[3]{x+1})^{3}-1\). This simplifies to \(f(g(x))=x+1-1\).

Step 2: Simplified form of \(f \circ g\)

After substitution, the expression simplifies down to \(f(g(x))=x\). Hence, the composite function \(f \circ g\) is \(x\).

Step 3: Function Composition \(g \circ f\)

Similarly, the composition of the functions \(g\) and \(f\) is denoted as \(g \circ f\). It is the function obtained by performing \(g\) after \(f\) on an input \(x\). In other words, it means substitute \(f(x)\) into \(g(x)\). The formula for function composition is: \(g(f(x))\). Here, substitute \(f(x)=x^{3}-1\) into \(g(x)\), thus \(g(f(x))=\sqrt[3]{(x^{3}-1)+1}\). This simplifies to \(g(f(x))=\sqrt[3]{x^{3}}\).

Step 4: Simplified form of \(g \circ f\)

After substitution, the expression simplifies down to \(g(f(x))=x\). Hence, the composite function \(g \circ f\) is \(x\).

Key Concepts

Composite Function

When we talk about composite functions, we're referring to the process of combining two functions to create a new one. A composite function like \(f \circ g\) means you will apply the function \(g\) first, and then \(f\) to the result of \(g\). Here's the basic idea:

• The notation \(f \circ g\) stands for \(f(g(x))\), meaning "apply \(g\) and then \(f\)".
• To compute \(f \circ g\), substitute the result from \(g(x)\) into \(f\).
• For example, if \(g(x) = \sqrt[3]{x+1}\) and \(f(x) = x^3 - 1\), then substituting \(g(x)\) into \(f\) becomes \(f(g(x)) = (\sqrt[3]{x+1})^3 - 1\).
• This simplifies to \(x + 1 - 1 = x\), making the composite function simply \(x\).

The beauty of composite functions lies in how we can chain transformations, applying one function after another, to simplify or otherwise manipulate expressions.

Composition of Functions

The composition of functions is essentially about embedding one function inside another. It allows us to see how one function affects another in sequence. Here's how it works:

• You have two functions, \(f(x)\) and \(g(x)\).
• The composition, either \(f \circ g\) or \(g \circ f\), tells the sequence in which to apply these functions.
• \(f \circ g\) means you first apply \(g\), and then use its result as the input for \(f\).
• For instance, with \(f(x)=x^3-1\) and \(g(x)=\sqrt[3]{x+1}\), doing \(g \circ f\) involves putting \(f\) into \(g\): \(g(x^3-1)=\sqrt[3]{(x^3-1)+1}\).
• This simplifies to \(\sqrt[3]{x^3}\), which turns out to be just \(x\) as well.

Composition can also help in finding or proving identities, like verifying how functions interact, often leading to simpler or more useful forms.

Inverse Functions

Inverse functions are functions that "undo" each other. If you apply an inverse function to a function, you should get the original input back. Think of it as a way to "reverse" the effect of another function.

• For two functions \(f\) and \(g\), if \(f(g(x)) = x\) and \(g(f(x)) = x\), then \(f\) and \(g\) are considered inverses of each other.
• In our example, the compositions \(f \circ g\) and \(g \circ f\) both resulted in \(x\). This shows that each function serves as the inverse of the other.
• The notation for the inverse of a function \(f\) is \(f^{-1}\).
• To find the inverse of a function, you generally swap the roles of \(x\) and \(y\) in the equation \(y = f(x)\) and solve for \(x\).

Getting comfortable with inverse functions means appreciating how they can "cancel out" certain effects, making them extremely useful in algebra and calculus.