Inverse Functions: A Step-by-Step Guide

Hey math whizzes! Ever feel like functions are playing hide-and-seek with you? Well, today we're going to become master detectives and find the hidden identity of a function: its inverse! We'll be tackling a specific example: finding the inverse of $f(x)=\sqrt{1-x}+10$ for the domain $(-\infty, 1]$ and then figuring out the domain of this sneaky inverse function. So grab your magnifying glasses, and let's dive into the fascinating world of inverse functions!

Understanding Inverse Functions: The Ultimate Function Flip

Alright guys, let's get down to business. What exactly is an inverse function? Think of it like this: if a function $f$ takes an input $x$ and gives you an output $y$, its inverse function, denoted as $f^{-1}$, does the exact opposite. It takes that output $y$ and gives you back the original input $x$. It's like a undo button for functions! This whole concept is super important in mathematics, especially when you're dealing with equations and trying to solve for variables. You see, if you have a function that's transformed your original value, the inverse function is the key to getting that original value back. It's all about reversing the process. For an inverse function to exist, the original function needs to be one-to-one, meaning each output corresponds to only one unique input. Our function $f(x)=\sqrt{1-x}+10$ definitely fits this bill, and we'll see why as we go along. When we talk about the domain and range of a function and its inverse, there's a beautiful symmetry happening. The domain of the original function becomes the range of its inverse, and the range of the original function becomes the domain of its inverse. It’s like a perfect mirror image, just flipped! This relationship is crucial for understanding how inverse functions behave and how they can be used to solve complex problems. So, keep this domain-range swap in mind as we work through our example, because it's going to be our secret weapon later on!

Finding the Inverse of $f(x)=\sqrt{1-x}+10$ Step-by-Step

Now, let's get our hands dirty with our specific function: $f(x)=\sqrt{1-x}+10$, with the domain $(-\infty, 1]$. Our mission, should we choose to accept it, is to find $f^{-1}(x)$ and its domain. Here’s the game plan, broken down into easy-to-follow steps:

Step 1: Replace $f(x)$ with $y$

First things first, we need to make things a little easier to work with. So, let's replace $f(x)$ with $y$. Our function now looks like this:

$y = \sqrt{1-x} + 10$

This is just a cosmetic change, but it helps us visualize the input ($x$) and output ($y$) relationship more clearly. It's like renaming your variables so they're easier to manage in a complex operation. We're essentially setting up the equation that defines our function's behavior.

Step 2: Swap $x$ and $y$

This is the magic step where we start the process of finding the inverse. Remember how we said an inverse function reverses the input and output? Well, we're going to physically swap the roles of $x$ and $y$ in our equation. This represents the core idea of inversion – what was an output is now an input, and vice versa.

$x = \sqrt{1-y} + 10$

Don't let this step confuse you, guys. It's the fundamental principle behind finding any inverse function. We're now looking at the equation from the perspective of the inverse function. The $x$ on this side is what would be the output of $f$, and the $y$ is what was the input of $f$. This swap is the algebraic key to unlocking the inverse function's definition.

Step 3: Solve for $y$

Our next mission is to isolate $y$ in our new equation. This means we need to get $y$ all by itself on one side of the equation. We'll use our trusty algebraic tools to achieve this.

First, subtract 10 from both sides:

$x - 10 = \sqrt{1-y}$

Now, to get rid of that pesky square root, we need to square both sides of the equation:

$(x - 10)^2 = (\sqrt{1-y})^2$

This simplifies to:

$(x - 10)^2 = 1 - y$

We're so close! Now, we just need to get $y$ on its own. Subtract 1 from both sides:

$(x - 10)^2 - 1 = -y$

Finally, multiply both sides by -1 to make $y$ positive:

$-((x - 10)^2 - 1) = y$

Or, if you prefer to distribute that negative sign:

$y = -(x^2 - 20x + 100) + 1$

$y = -x^2 + 20x - 100 + 1$

$y = -x^2 + 20x - 99$

So, our inverse function, $f^{-1}(x)$, is:

$f^{-1}(x) = -x^2 + 20x - 99$

See? Not so scary when you break it down, right?

Determining the Domain of the Inverse Function ($f^{-1}(x)$)

Now for the other crucial part: finding the domain of $f^{-1}(x)$. Remember our earlier chat about the domain of $f$ becoming the range of $f^{-1}$, and the range of $f$ becoming the domain of $f^{-1}$? This is where that knowledge comes in handy!

Step 4: Find the Range of the Original Function ($f(x)$)

To find the domain of $f^{-1}(x)$, we first need to determine the range of our original function, $f(x) = \sqrt{1-x} + 10$. The domain of $f(x)$ is given as $(-\infty, 1]$. Let's think about the square root part, $\sqrt{1-x}$.

Since the square root function $\sqrt{\cdot}$ always produces non-negative values (i.e., $\ge 0$), the term $\sqrt{1-x}$ will always be greater than or equal to 0.

Now, consider the entire function $f(x) = \sqrt{1-x} + 10$. Because $\sqrt{1-x} \ge 0$, when we add 10 to it, the smallest value $f(x)$ can take is $0 + 10 = 10$. As $x$ approaches negative infinity, $1-x$ becomes increasingly large, and so does $\sqrt{1-x}$. Therefore, $f(x)$ can take any value greater than or equal to 10.

So, the range of $f(x)$ is $[10, \infty)$.

Step 5: The Domain of $f^{-1}(x)$ is the Range of $f(x)$

Here's the beautiful part: the domain of the inverse function $f^{-1}(x)$ is precisely the range of the original function $f(x)$.

Since the range of $f(x)$ is $[10, \infty)$, the domain of $f^{-1}(x)$ is also $[10, \infty)$.

In interval notation, this is written as [10, $\infty$).

Why This Domain Restriction is Crucial

Let's think about why this domain restriction is so important for our inverse function $f^{-1}(x) = -x^2 + 20x - 99$. If we didn't have this restriction, the function $f^{-1}(x)$ would be a parabola opening downwards, which isn't one-to-one (meaning it fails the horizontal line test and wouldn't have a true inverse). By restricting the domain of $f^{-1}(x)$ to $[10, \infty)$, we are essentially selecting only the portion of the parabola that corresponds to the original function's behavior. This ensures that $f^{-1}(x)$ is indeed the correct inverse of $f(x)$ and that it passes the vertical line test, which is a requirement for any function.

This domain restriction is a direct consequence of the original function's domain and the nature of the square root operation. When we performed the algebraic steps to find the inverse, we squared both sides. Squaring can introduce extraneous solutions, but by understanding the range of the original function, we can correctly identify the valid domain for the inverse. It’s a way to ensure our mathematical operations are consistent and that we're not creating something that doesn’t accurately reflect the original relationship.

Final Check: Does it all add up?

So, to recap, we found that:

• The inverse function is $f^{-1}(x) = -x^2 + 20x - 99$.
• The domain of $f^{-1}(x)$ is $[10, \infty)$.

It's always a good idea to do a quick check. Let's pick a value from the domain of $f(x)$, say $x=0$.

$f(0) = \sqrt{1-0} + 10 = \sqrt{1} + 10 = 1 + 10 = 11$.

Now, let's plug this output, $11$, into our inverse function $f^{-1}(x)$:

$f^{-1}(11) = -(11)^2 + 20(11) - 99 = -121 + 220 - 99 = 99 - 99 = 0$.

And voilà! We got our original input, $0$, back. This confirms that our inverse function and its domain are correct. It’s like finding a matching pair of socks; everything fits perfectly!

Conclusion: Mastering Inverse Functions

Finding inverse functions and their domains might seem a bit daunting at first, but as you've seen, it's all about following a systematic process. By understanding the relationship between a function and its inverse, and by carefully performing algebraic manipulations, you can conquer any inverse function problem that comes your way. Remember the key steps: swap $x$ and $y$, solve for $y$, and then use the range of the original function to determine the domain of the inverse. Keep practicing, and you'll be an inverse function pro in no time! This skill is fundamental not just in algebra, but also in calculus and beyond, opening doors to solving more complex mathematical puzzles. So, keep exploring, keep questioning, and keep those mathematical gears turning, guys!