Problem 30 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

(a) \(f \circ g = x\), (b) \(g \circ f = x\).

Step 1: Find \(f \circ g\)

Replace \(x\) in \(f(x)\) with \(g(x)\). Therefore, \(f \circ g = f(g(x))= \sqrt[3]{g(x)-1}= \sqrt[3]{(x^{3}+1)-1}\)}, which simplifies to \(\sqrt[3]{x^{3}} = x\).

Step 2: Find \(g \circ f\)

Replace \(x\) in \(g(x)\) with \(f(x)\). Therefore, \(g \circ f = g(f(x))= (f(x))^{3}+1)= (\sqrt[3]{x-1})^{3}+1\), which simplifies to \(x-1+1=x\).

Key Concepts

Inverse Functions

Inverse functions essentially go in reverse. They "undo" what the original function does. For a pair of functions, like the ones given:

• \(f(x) = \sqrt[3]{x-1}\)
• \(g(x) = x^{3} + 1\)

To find if they are inverses, we evaluate the compositions \(f \circ g\) and \(g \circ f\). If both expressions simplify back to \(x\), the functions are true inverses of each other.

In the example, both compositions simplify correctly:

• \(f(g(x)) = x\)
• \(g(f(x)) = x\)

Thus, confirming \(f(x)\) and \(g(x)\) are inverses. In inverse functions, any element that goes through one function and its inverse gets returned to its original state.

Cubic Functions

Cubic functions are a type of polynomial function characterized by the highest degree term being cubed (to the power of 3). An example of a cubic function is

• \(g(x) = x^{3} + 1\)

This function shapes like a stretched-out "S" when graphed. It has curves due to the cubic term dominating behavior at extreme values of \(x\).

Critical points of cubic functions reveal interesting behavior like turning points or inflections. They don't necessarily imply min or max points which is distinctive of their curve.

Cubic functions can be reversed when a corresponding radical function partners up with them. This is because the cubic can be effectively neutralized through root extraction, demonstrated in their inverse relationship.

Radical Functions

Radical functions include roots—square roots, cube roots, and higher ones—which determines their nature. One such function, used in the exercise, is the cube root function:

• \(f(x) = \sqrt[3]{x-1}\)

The cube root operation takes a number times itself three times to retrieve the original number from its cube. They’re unique since odd roots (like cube roots) can take negatives and yield negative results, as opposed to even roots.

In operations, combining cube roots with cubic terms often leads to simplifications. This is because they counteract each other thanks to their mathematical relationship. For instance, \(\sqrt[3]{x^3} = x\).

Radical functions, like \(f(x)\), in this problem particularly show how radical functions work with polynomials to return linear relationships, when paired just right.