Problem 52 - Elementary and Intermediate Algebra

Question

Each of the following functions is one-to-one. Find the inverse of each function and express it using \(f^{-1}(x)\) notation. \(f(x)=\frac{3}{x^{3}}-1\)

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \sqrt[3]{\frac{3}{x + 1}}\).

Step 1: Replace f(x) with y

Start by writing the function with public isc as a variable name in place of f(x): \( y = \frac{3}{x^3} - 1 \)

Step 2: Swap x and y

To find the inverse of the function, swap the x and y variables: \( x = \frac{3}{y^3} - 1 \)

Step 3: Solve for y

Our goal is to express y in terms of x. Add 1 to both sides to isolate the term with y: \( x + 1 = \frac{3}{y^3} \).Next, multiply both sides by \( y^3 \): \( (x + 1) y^3 = 3 \).

Step 4: Isolate y^3

Divide both sides by \(x + 1\) to isolate \(y^3\): \( y^3 = \frac{3}{x + 1} \).

Step 5: Take the cube root

To solve for y, take the cube root of both sides: \( y = \sqrt[3]{\frac{3}{x + 1}} \). This is the expression for the inverse in terms of y.

Step 6: Express the inverse function in proper notation

Substitute f^(-1) (x) for y to express the inverse function properly: \( f^{-1}(x) = \sqrt[3]{\frac{3}{x + 1}} \)

Key Concepts

One-to-One Functions

One-to-one functions are a special category of functions where each element in the function's domain is paired with a unique element in the codomain. This means that no two different inputs produce the same output. An easy way to test if a function is one-to-one is by using the Horizontal Line Test. If any horizontal line intersects the graph of the function at most once, the function is one-to-one.

Understanding one-to-one functions is crucial. It's because invertible functions, which have inverse functions, must be one-to-one. Only then can we "reverse" the process to find unique outputs for each input from the codomain. When we speak of the inverse, we mean that if the original function pairs some input with an output, the inverse will pair that output back to the input it originated from.

For example, if function \( f \) maps \( 2 \to 5 \), then the inverse function \( f^{-1} \) will map \( 5 \to 2 \). This unique pairing allows us to confidently "go backwards" from any given result to the input.

Algebraic Functions

Algebraic functions are mathematical expressions formed using arithmetic operations like addition, subtraction, multiplication, division, and exponentiation. They typically involve variables like \(x\) and correspond to polynomial, rational, and radical expressions. Functions such as \(f(x) = x^2 + 3x + 2\), \(g(x) = \frac{3}{x}\), and \(h(x) = \sqrt{x}\) are all examples of algebraic functions.

These types of functions allow us to model and analyze various situations in mathematics. They can systematically describe relationships and patterns, much like our given function \(f(x) = \frac{3}{x^3} - 1\), which utilizes a rational expression in its structure. When dealing with algebraic functions, understanding the position of terms and operations is vital. This understanding assists in manipulating and solving equations, such as when we determine the inverse function.

In the context of inverses, transforming the algebraic function \(f(x)\) into \(f^{-1}(x)\) often involves swapping the roles of the dependent and independent variables and then solving the resulting expression. This step is a central part of finding inverse functions, especially for complex expressions.

Cube Root

The cube root operation is the inverse of cubing a number. When we say "cube root", we mean a value that, when raised to the power of three, yields the original number. For example, the cube root of 27 is 3 because \(3^3 = 27\). We use the notation \(\sqrt[3]{x}\) for the cube root.

Understanding cube roots is essential in cases where variables are fixed with an exponent of three. In this scenario, we often come across equations that require us to isolate a term that is cubed. To "undo" the cubing and solve for the variable, we apply the cube root.

In our example, when finding the inverse function \(f^{-1}(x) = \sqrt[3]{\frac{3}{x+1}}\), taking the cube root is crucial. This step reverses the cubing effect from the original expression. By applying the cube root to \(\frac{3}{x+1}\), we effectively find a value for \(y\) that satisfies the condition of the inverse function. This manipulation is pivotal in algebraic transformations when one needs to isolate variables in expressions involving cube powers.