Problem 25 - Algebra Form and Function

Question

Check that the functions are inverses. $$ f(x)=1+7 x^{3} \text { and } g(t)=\sqrt[3]{\frac{t-1}{7}} $$

Step-by-Step Solution

Answer

Answer: Yes, the functions are inverses of each other.

Step 1: Perform f(g(t))

We'll substitute g(t) into f(x) to get f(g(t)): $$ f(g(t)) = 1 + 7 \left(\sqrt[3]{\frac{t-1}{7}}\right)^3 $$

Step 2: Simplify f(g(t))

Now, we'll simplify the expression: $$ f(g(t)) = 1 + 7 \cdot \frac{t-1}{7} $$ $$ f(g(t)) = 1 + (t-1) $$ $$ f(g(t)) = t $$

Step 3: Perform g(f(x))

We'll substitute f(x) into g(t) to get g(f(x)): $$ g(f(x)) = \sqrt[3]{\frac{(1+7x^3)-1}{7}} $$

Step 4: Simplify g(f(x))

We'll simplify the expression: $$ g(f(x)) = \sqrt[3]{\frac{7x^3}{7}} $$ $$ g(f(x)) = \sqrt[3]{x^3} $$ $$ g(f(x)) = x $$ Since both f(g(t)) = t and g(f(x)) = x are true, the given functions are inverses of each other.

Key Concepts

Composition of Functions

The composition of functions involves taking one function and applying it to the result of another function, which allows us to see how the functions interact with each other.
In this exercise, we are using two functions: \( f(x) = 1 + 7x^3 \) and \( g(t) = \sqrt[3]{\frac{t-1}{7}} \).
When we perform the composition \( f(g(t)) \), it means we plug \( g(t) \) into \( f(x) \). This is mathematically written as \( f(g(t)) = 1 + 7\left(\sqrt[3]{\frac{t-1}{7}}\right)^3 \).
After simplifying, it becomes clear that \( f(g(t)) = t \), indicating that applying \( f \) after \( g \) returns us to the original input \( t \).Similarly, we can perform \( g(f(x)) \), where we substitute \( f(x) \) into \( g(t) \).
This is written as \( g(f(x)) = \sqrt[3]{\frac{(1+7x^3)-1}{7}} \). By simplifying this expression, it leads to \( g(f(x)) = x \), which shows the reverse composition returns the original input \( x \).
Therefore, the functions effectively cancel each other out, proving they are inverses.

Cube Roots

Cube roots play a critical role in finding inverse functions in this problem. The cube root of a number \( a \) is a value \( b \) such that \( b^3 = a \). When expressed as a function, it can be noted as \( \sqrt[3]{a} \).In function \( g(t) = \sqrt[3]{\frac{t-1}{7}} \), the cube root is used to effectively "undo" an operation in another function.
From the original step-by-step solution, when verifying the inverse relationship with \( g(f(x)) \), simplifying gives \( g(f(x)) = \sqrt[3]{x^3} \), which simplifies to \( x \).
This operation showcases how cube roots are essential for simplifying expressions involving cubic terms, supporting the simplification needed to find the inverse of cubic functions.
The understanding of cube roots is crucial as it's the mechanism used to determine the inverse relationships in functions involving cubic terms.

Function Simplification

Simplification is a powerful step to verify reverse operations and to confirm functions as inverses. When functions are composed, simplification reduces complex expressions.
In this exercise, simplifying involves resolving cube roots and canceling terms, ultimately revealing the input value of a function.For instance, when simplifying \( f(g(t)) \), we start with \( 1 + 7\left(\frac{t-1}{7}\right) \).
This reduces by canceling out the 7 to \( 1 + (t - 1) \), which further simplifies to \( t \), indicating that the operations within the function have been resolved.
Similarly, with \( g(f(x)) = \sqrt[3]{\frac{7x^3}{7}} \), simplifying gives \( \sqrt[3]{x^3} \) which results in \( x \), confirming the story of perfect cancellation.
These simplifications are essential in demonstrating how functions reverse each other, hence validating them as inverses.
The process of simplification allows one to directly see the growth or shrinkage factors, such as factorizations, and how they affect the result due to their powerful canceling effects.