Domain Restriction: Inverting Non-Invertible Functions

Let's dive into the fascinating world of functions and their inverses! Specifically, we're going to explore how restricting the domain of a non-invertible function allows us to create an inverse. We'll use the example of $f(x) = (x-3)^2 + 2$ to illustrate this concept. So, grab your thinking caps, and let's get started!

Understanding Non-Invertible Functions

Before we jump into domain restrictions, it's crucial to understand what makes a function non-invertible in the first place. A function is invertible if it's one-to-one, also known as injective. This means that for every output (y-value), there is only one unique input (x-value). Graphically, a one-to-one function passes the horizontal line test: any horizontal line drawn will intersect the graph at most once.

The function $f(x) = (x-3)^2 + 2$ is a quadratic function, which, when plotted, forms a parabola. Parabolas notoriously fail the horizontal line test. Think about it: for almost any y-value above the vertex of the parabola, there will be two corresponding x-values. This is because the parabola is symmetric around its vertex. Because of this symmetry, the original function, without any domain restrictions, isn't invertible. We can't simply find a single inverse function that works for all possible x-values.

Why is invertibility so important? Well, the inverse function, denoted as $f^{-1}(x)$, essentially "undoes" what the original function does. If $f(a) = b$, then $f^{-1}(b) = a$. For this to work reliably, we need a clear, unambiguous relationship between inputs and outputs, which is what the one-to-one property guarantees. When a function isn't one-to-one, trying to define an inverse leads to ambiguity. For example, if $f(x)$ gives the same output for two different $x$ values, which $x$ should the inverse return when given that output? This is why we need to get a little clever and restrict the domain.

The Power of Domain Restriction

So, how do we make a non-invertible function invertible? The trick is to restrict its domain. By carefully choosing a subset of the original domain, we can create a new function that is one-to-one. This is where the magic happens! For our example, $f(x) = (x-3)^2 + 2$, the vertex of the parabola is at the point (3, 2). The parabola is symmetric around the vertical line x = 3. This means that for any x-value to the left of 3, there's a corresponding x-value to the right of 3 that produces the same y-value.

To make the function one-to-one, we can chop off one half of the parabola. The problem statement specifies that we are restricting the domain to $[3, "), which means we are only considering x-values greater than or equal to 3. This effectively eliminates the left half of the parabola. The remaining portion does pass the horizontal line test. For every y-value greater than or equal to 2, there is now only one corresponding x-value in the restricted domain. Therefore, with this restriction, the function becomes invertible.

Restricting the domain is a powerful technique. It allows us to work with inverses even when the original function isn't invertible over its entire domain. It's like saying, "Okay, I know this function isn't invertible in general, but if I only look at this specific part of it, then it is invertible." This is a common practice in mathematics and is essential for defining inverses for functions like trigonometric functions as well.

Finding the Inverse Function

Now that we've restricted the domain of $f(x)$ to $[3, "), let's find its inverse. Here's how we do it:

1. Replace $f(x)$ with $y$: So, we have $y = (x-3)^2 + 2$.

2. Swap $x$ and $y$: This gives us $x = (y-3)^2 + 2$.

3. Solve for $y$: This is where we isolate $y$ on one side of the equation.

   • Subtract 2 from both sides: $x - 2 = (y-3)^2$
   • Take the square root of both sides: $\sqrt{x-2} = y - 3$. Important: Since we restricted the domain to $x ">= 3$, we only consider the positive square root. If we had restricted the domain to $x \leq 3$, we would take the negative square root.
   • Add 3 to both sides: $y = \sqrt{x-2} + 3$

4. Replace $y$ with $f^{-1}(x)$: This gives us the inverse function: $f^{-1}(x) = \sqrt{x-2} + 3$.

So, the inverse of the function $f(x) = (x-3)^2 + 2$ with the domain restricted to $[3, ") is indeed $f^{-1}(x) = \sqrt{x-2} + 3$. This is a crucial step, as it allows us to reverse the operation of the original function within the specified domain.

Determining the Domain of the Inverse Function

The domain of the inverse function is the range of the original function (with the restricted domain). Let's think about why this is so important. The inverse function essentially "undoes" the original function. Therefore, the inputs of the inverse function must be the outputs of the original function. If the original function never produces a particular output value, then that value cannot be a valid input for the inverse function.

For $f(x) = (x-3)^2 + 2$ with the domain $[3, "), the smallest possible value of $(x-3)^2$ is 0, which occurs when $x = 3$. Therefore, the smallest possible value of $f(x)$ is $0 + 2 = 2$. As $x$ increases from 3, $(x-3)^2$ also increases, and so does $f(x)$. There's no upper bound on $x$, so there's no upper bound on $f(x)$. This means the range of $f(x)$ is $[2, ").

Therefore, the domain of the inverse function, $f^{-1}(x) = \sqrt{x-2} + 3$, is $[2, "). We can also see this directly from the inverse function. The expression inside the square root, $x-2$, must be non-negative. So, $x-2 \geq 0$, which means $x \geq 2$. This confirms that the domain of the inverse function is $[2, ").

Understanding the relationship between the domain of the original function and the range, and therefore the domain of the inverse, is essential for working with inverse functions. It ensures that the inverse function is properly defined and that we are using it within its valid input range. Always remember to check the domain! It can save you from making mistakes.

Putting it All Together

So, to summarize:

• The function $f(x) = (x-3)^2 + 2$ is not invertible over its entire domain.
• By restricting the domain to $[3, "), we make the function one-to-one and therefore invertible.
• The inverse function is $f^{-1}(x) = \sqrt{x-2} + 3$.
• The domain of the inverse function is $[2, ").

This example demonstrates how domain restriction is a crucial technique for finding inverses of non-invertible functions. By carefully choosing the domain, we can create a one-to-one function and find its inverse, opening up a whole new world of mathematical possibilities. Remember, the key is to make the original function one-to-one by restricting its domain appropriately. Keep practicing, and you'll master this technique in no time! You got this, guys!