Question

The function \(f\) is one-to-one. (a) Find its inverse function \(f^{-1}\) and check your answer. (b) Find the domain and the range of \(f\) and \(f^{-1}\). $$ f(x)=\frac{1}{9}(x-1)^{2}+2, \quad x \geq 1 $$

Short answer

The inverse function is \( f^{-1}(y) = 1 + 3\sqrt{y - 2} \). The domain of \( f \) is \([1, \infty)\) and the range is \([2, \infty)\). The domain of \( f^{-1} \) is \([2, \infty)\) and the range is \([1, \infty)\).

Step-by-step solution

- Define the Inverse Function

Start by replacing the function notation with a variable. Let \[ y = \frac{1}{9}(x-1)^{2}+2 \] Now, solve for \(x\) in terms of \(y\).

- Isolate the Variable

Subtract 2 from both sides to isolate the quadratic term: \[ y - 2 = \frac{1}{9}(x-1)^2 \].

- Eliminate the Fraction

Multiply both sides by 9 to get rid of the fraction: \[ 9(y - 2) = (x-1)^2 \].

- Solve for the Square Root

Take the square root of both sides. Since \(x \geq 1\), use the positive root: \[ \sqrt{9(y - 2)} = x - 1 \].

- Solve for \(x\)

Add 1 to both sides to isolate \(x\): \[ x = 1 + \sqrt{9(y - 2)} \] This represents the inverse function: \[ f^{-1}(y) = 1 + 3\sqrt{y - 2} \].

- Verify the Inverse

To verify, compose \(f\) and its inverse: Let's find \(f(f^{-1}(x))\). Substitute \(f^{-1}(x)\) into \(f\): \[ f(f^{-1}(x)) = f(1 + 3\sqrt{x-2}) = \frac{1}{9}((1 + 3\sqrt{x-2}) - 1)^2 + 2 \]. Simplify the expression: \[ \frac{1}{9}(3\sqrt{x-2})^2 + 2 = \frac{1}{9}*9(x-2) + 2 = x-2 + 2 = x \]. So, \( f(f^{-1}(x)) = x \). Thus, the inverse function is correct.

- Find the Domain and Range

For the original function \(f(x)\): - Domain: Given as \(x \geq 1\). - Range: Since \(f(x)\) adds 2 to a squared term, \(f(x)\) values start at 2 and go to infinity, i.e., \([2, \infty)\). For the inverse function \(f^{-1}(x)\): - Domain: Corresponds to the range of \(f(x)\), i.e., \([2, \infty)\). - Range: Corresponds to the domain of \(f(x)\), i.e., \([1, \infty)\).

Key concepts

One-to-One Functions

One-to-one functions (also called injective functions) are functions where each element of the range is mapped to by exactly one element of the domain. This ensures that the function passes the horizontal line test - no horizontal line intersects the graph of the function more than once. Here’s why this is important:

• If a function is one-to-one, it means we can find its inverse. The inverse function reverses the role of inputs and outputs, giving us the original input from the original output.
• This determination is essential for finding the inverse function, as demonstrated by the function in the exercise, where we identify one-to-one behavior in order to move forward with finding its inverse.
• In practical terms, this property simplifies data processing in fields where unique pairings are required, such as cryptography and data integrity checks.

So, for the function provided, ensuring it's one-to-one allows us to solve for and verify the inverse function confidently.

Domain and Range

Understanding the domain and range of a function helps us determine where the function is defined and its output values.

• The domain of a function is the set of all possible input values (x-values) that the function can accept.
• The range of a function is the set of all possible output values (y-values) that the function can produce.

In our example, we have an original function:
\(f(x) = \frac{1}{9}(x-1)^2 + 2\), with a given domain of \(x \geq 1\). This is because the function includes a squared term that naturally restricts x-values to those greater or equal to 1. The function also shifts upwards due to the +2, ensuring the smallest possible value of the output (range) starts from 2. This range is noted as [2, \infty).
Similarly, for the inverse function \(f^{-1}(x) = 1 + 3\sqrt{x - 2}\), the roles of domain and range are reversed.

The domain corresponds to the range of the original function, meaning it starts at 2 and extends to infinity.

Conversely, the range reflects the domain of the original function, starting at 1 and also extending to infinity.

Accurate specifications of domain and range are vital in mathematical functions and their applications, ensuring valid inversions and interpretations.

Function Composition

Function composition involves applying one function to the results of another. It's like plugging the output from one function directly into another function.
For our case, verifying the inverse function essentially uses function composition:

• We take the inverse function \(f^{-1}(x)\) and substitute it into the original function \(f(x)\), and vice versa, to check if the compositions return the initial input values. This ensures they are true inverses of each other.
• In our exercise, we composed \(f(f^{-1}(x))\): \(f\left(1 + 3\sqrt{x-2}\right) = x\). This composition simplifies through proper substitution and verification steps, showing the input returned unchanged.

Function composition is a fundamental concept for understanding inverse functions. It also broadens one's toolset for working with complex mathematical models, including nested functions and multi-step calculations. By mastering composition, students can ensure accuracy and depth in their problem-solving approaches.