Inverse Function Of F(x) = Sqrt(x) - 8: A Full Guide

Hey math whizzes! Today, we're diving deep into the awesome world of inverse functions. We're going to tackle a specific problem: finding the inverse function and its domain for $f(x) = \sqrt{x} - 8$. This might sound a bit tricky, but trust me, by the end of this article, you'll be an inverse function pro! We'll break down each step, explain the why behind it, and make sure you totally get it. So, grab your calculators, your favorite study snacks, and let's get this math party started!

Understanding Inverse Functions: The Basics

First off, what exactly is an inverse function? Think of it as the "undoing" function. If a function $f$ takes an input $x$ and gives you an output $y$, its inverse function, denoted as $f^{-1}$, takes that output $y$ and gives you back the original input $x$. It's like a secret code – the function encrypts it, and the inverse function decrypts it. For an inverse function to exist, the original function must be one-to-one, meaning each output corresponds to only one input. Our function, $f(x) = \sqrt{x} - 8$, involves a square root, which, for its principal value, is one-to-one, so we're good to go!

Step-by-Step: Finding the Inverse Function $f^{-1}(x)$

Alright guys, let's get down to business. Finding the inverse function $f^{-1}(x)$ for $f(x) = \sqrt{x} - 8$ involves a few key steps. It's like following a recipe – stick to the instructions, and you'll get a delicious mathematical result! We'll be using the properties of functions and their inverses to guide us through this. This process is fundamental in algebra and calculus, so mastering it will open up a lot of doors for you in your math journey. Don't just memorize the steps; try to understand the logic behind each one. This deep understanding will help you solve even more complex problems down the line. So, let's start with the first, crucial step in our inverse function quest.

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

The very first thing we do is rewrite our function $f(x) = \sqrt{x} - 8$ by simply replacing $f(x)$ with the variable $y$. This makes it easier to manipulate algebraically. So, our equation becomes:

$$y = \sqrt{x} - 8$$

This substitution is purely for convenience; it doesn't change the function's behavior or its relationship between inputs and outputs. It's just a common convention in mathematics to use $y$ when representing the output of a function. Think of it as setting up the stage for the next act in our algebraic drama. This simple change allows us to treat the equation more like a standard algebraic equation, which is essential for the next steps in finding the inverse. So, whenever you see $f(x)$, just swap it out for $y$ to get the ball rolling on your inverse function calculations. It’s a small step, but a really important one!

Step 2: Swap $x$ and $y$

Now for the magic step that defines the inverse! To find the inverse function, we literally swap the roles of $x$ and $y$. Remember, the inverse function takes the output of the original function and turns it back into the input. So, we exchange every $x$ for a $y$ and every $y$ for an $x$ in our equation.

$$x = \sqrt{y} - 8$$

This step is the core of finding an inverse. By swapping $x$ and $y$, we are essentially saying, "Okay, if the original function maps $x$ to $y$, then the inverse function must map $y$ back to $x$." This is why the roles are interchanged. It's the conceptual leap that makes the inverse function work. This maneuver might seem a bit arbitrary at first, but it's mathematically sound and directly leads us to the equation representing the inverse relationship. Keep this step in mind, as it's the linchpin of the entire process. We're getting closer to our final answer, so let's keep pushing!

Step 3: Solve for $y$

Our next mission, should we choose to accept it, is to isolate $y$ in the equation we got from Step 2: $x = \sqrt{y} - 8$. This is where our algebraic skills really shine. We need to get $y$ all by itself on one side of the equation.

First, add 8 to both sides to isolate the square root term:

$$x + 8 = \sqrt{y}$$

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

$$(x + 8)^2 = (\sqrt{y})^2$$

$$(x + 8)^2 = y$$

And there we have it! We've successfully solved for $y$. This $y$ is our inverse function!

Step 4: Replace $y$ with $f^{-1}(x)$

The final touch in naming our inverse function is to replace $y$ with the proper notation, $f^{-1}(x)$. So, our inverse function is:

$$f^{-1}(x) = (x + 8)^2$$

We've successfully found the algebraic expression for the inverse function. High five, everyone! We navigated through the algebraic manipulations, and the result is a neat and tidy equation. Remember, this $f^{-1}(x)$ is the function that "undoes" what $f(x)$ did. If you plug a value into $f(x)$ and then plug the result into $f^{-1}(x)$, you should get your original value back. It's a testament to the elegance of mathematics that we can reverse these operations to find such corresponding functions. This inverse function relationship is super important in many areas of math, so be proud of mastering this step!

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

Now, just finding the formula for the inverse function isn't enough, guys. We also need to figure out its domain. The domain of the inverse function $f^{-1}(x)$ is actually related to the range of the original function $f(x)$. This is a crucial connection to remember!

Let's first consider the original function $f(x) = \sqrt{x} - 8$. The square root function, $\sqrt{x}$, is defined only for non-negative values of $x$. So, the domain of $f(x)$ is $x \geq 0$.

Now, what about the range of $f(x)$? Since $\sqrt{x}$ always outputs a non-negative number (i.e., $\sqrt{x} \geq 0$), subtracting 8 from it means the output $f(x)$ will always be greater than or equal to -8. So, the range of $f(x)$ is $f(x) \geq -8$.

Key Takeaway: The domain of $f^{-1}(x)$ is the range of $f(x)$.

Since the range of $f(x)$ is $f(x) \geq -8$, the domain of our inverse function $f^{-1}(x)$ must also be $x \geq -8$. This means that we can only plug values into $f^{-1}(x)$ that are greater than or equal to -8.

So, our complete answer, including the inverse function and its domain, is:

f^{-1}(x) = (x + 8)^2 \quad \text{with domain } x \geq -8$. This domain restriction is super important. Without it, the inverse function $(x+8)^2$ would behave differently than intended, specifically for negative inputs. The original function $f(x)=\sqrt{x}-8$ restricted its outputs to be $-8$ or greater. Therefore, its inverse must restrict its inputs to be $-8$ or greater to correctly "undo" the original function. This connection between the domain of the inverse and the range of the original function is a fundamental concept in understanding inverse functions and their properties. It ensures that the inverse operation perfectly reverses the original one, maintaining the integrity of the input-output relationship. ## Verification: Checking Our Work! It's always a good idea to double-check our work, right? We can verify if $f^{-1}(x) = (x + 8)^2$ with $x \geq -8$ is indeed the inverse of $f(x) = \sqrt{x} - 8$ by checking two conditions: 1. $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}(x)$. 2. $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f(x)$. Let's test the first condition: $f(f^{-1}(x)) = f((x + 8)^2) = \sqrt{(x + 8)^2} - 8

Since the domain of $f^{-1}(x)$ is $x \geq -8$, this means $x+8 \geq 0$. Therefore, $\sqrt{(x + 8)^2} = x + 8$.

$$f(f^{-1}(x)) = (x + 8) - 8 = x$$

This condition holds true for $x \geq -8$!

Now, let's test the second condition:

$$f^{-1}(f(x)) = f^{-1}(\sqrt{x} - 8) = ((\sqrt{x} - 8) + 8)^2$$

The domain of $f(x)$ is $x \geq 0$, which means $\sqrt{x}$ is defined.

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

This condition holds true for $x \geq 0$!

Both conditions are met, so we can be super confident that our inverse function and its domain are correct. This verification process is like a final quality check. It confirms that the function we found truly reverses the operation of the original function, ensuring that when we go from $x$ to $f(x)$ and then to $f^{-1}(f(x))$, we end up right back where we started. This symmetry and reversibility are key properties of inverse functions and are fundamental to many mathematical concepts.

Comparing with the Options

Now, let's look at the options provided:

A. $f^{-1}(x)=x2+8 ; x geq 0$ B. $f^{-1}(x)=x2+8 ; x geq-8$ C. $f^{{-1}(x)=(x+8)}2 ; x geq 0$ D. $f^{{-1}(x)=(x+8)}2 ; x geq-8$

Based on our detailed step-by-step derivation and verification, our derived inverse function is $f^{-1}(x) = (x + 8)^2$ and its domain is $x \geq -8$. This perfectly matches Option D.

It's important to note why the other options are incorrect:

• Options A and B have the wrong form for the inverse function ($x^2+8$ instead of $(x+8)^2$). This arises from incorrectly performing the algebraic steps, possibly by not properly isolating the square root before squaring, or by misapplying the inverse relationship.
• Option C has the correct form of the inverse function ($f^{-1}(x) = (x+8)^2$) but the incorrect domain ($x \geq 0$). This mistake often happens when someone assumes the domain of the inverse is the same as the domain of the original function, instead of remembering that the domain of the inverse is the range of the original function.

Understanding why the incorrect options are wrong is just as important as knowing the right answer. It solidifies your grasp of the concepts and helps you avoid similar pitfalls in the future. Each incorrect option highlights a common misunderstanding about inverse functions, their algebraic manipulation, or domain/range relationships.

Conclusion: You've Got This!

So there you have it, guys! We've successfully found the inverse function and its domain for $f(x) = \sqrt{x} - 8$. We learned that the inverse function is $f^{-1}(x) = (x + 8)^2$, and critically, its domain is restricted to $x \geq -8$. This domain restriction is directly tied to the range of the original function, which is a vital concept when working with inverse functions.

Remember, the process involves replacing $f(x)$ with $y$, swapping $x$ and $y$, solving for $y$, and then replacing $y$ with $f^{-1}(x)$. Don't forget the crucial step of determining the domain of the inverse by looking at the range of the original function. Math can be like solving puzzles, and each step brings you closer to the complete picture.

Keep practicing, keep questioning, and never be afraid to dive deeper into the fascinating world of mathematics. You've got the tools and the knowledge now to tackle similar problems with confidence. Keep up the amazing work, and happy solving!