Finding The Inverse Of F(x) = √(x-5): A Step-by-Step Guide
Hey guys! Today, we're diving into the fascinating world of inverse functions, specifically focusing on how to find the inverse of the function f(x) = √(x-5). This is a common type of problem in mathematics, and mastering it will definitely help you in your calculus and algebra endeavors. So, let's break it down step-by-step, making it super easy to understand. We'll walk through the theory, the process, and ultimately arrive at the correct answer together.
Understanding Inverse Functions
Before we jump into solving for the inverse of f(x) = √(x-5), it’s crucial to understand what an inverse function actually is. Think of a function like a machine: you put something in (the input, usually x), and the machine spits out something else (the output, usually f(x) or y). An inverse function is like a machine that does the opposite; it takes the output and returns the original input. Mathematically speaking, if f(a) = b, then the inverse function, denoted as f⁻¹(x), should satisfy f⁻¹(b) = a.
In simpler terms, the inverse function “undoes” what the original function does. However, not every function has an inverse. For a function to have an inverse, it must be a one-to-one function. A one-to-one function means that each input (x) corresponds to exactly one output (y), and each output (y) corresponds to exactly one input (x). Graphically, a function is one-to-one if it passes the horizontal line test, meaning that no horizontal line intersects the graph of the function more than once. This one-to-one correspondence is the cornerstone of invertibility.
For our function, f(x) = √(x-5), we need to consider its domain. The square root function is only defined for non-negative values, so x - 5 ≥ 0, which means x ≥ 5. This restriction on the domain will be important later when we consider the range of the inverse function. The original function will only produce non-negative values since the square root of a number is always non-negative. This means the range of f(x) is y ≥ 0. Understanding these domain and range restrictions is vital in finding and correctly stating the inverse function.
Steps to Find the Inverse Function
Okay, now that we've got the basics down, let's tackle the actual process of finding the inverse. There's a simple, straightforward method we can follow, usually involving these steps:
- Replace f(x) with y: This is simply a notational change to make the algebra a bit easier to handle. So, we rewrite f(x) = √(x-5) as y = √(x-5).
- Swap x and y: This is the key step in finding the inverse. We're essentially reversing the roles of input and output. After swapping, our equation becomes x = √(y-5).
- Solve for y: Now we need to isolate y on one side of the equation. This usually involves algebraic manipulation. To get rid of the square root, we can square both sides of the equation: x² = (√(y-5))², which simplifies to x² = y - 5. Next, we add 5 to both sides to isolate y: x² + 5 = y. This gives us a potential inverse function.
- Replace y with f⁻¹(x): This is the final notational step. We replace y with the inverse function notation f⁻¹(x). So, our potential inverse function is f⁻¹(x) = x² + 5.
Verifying the Inverse and Considering Domain Restrictions
We've arrived at a potential inverse function, but we're not quite done yet! It's super important to verify that the function we found is indeed the inverse. To do this, we can check if f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. This essentially means we're plugging the inverse function into the original function and vice versa, and checking if we get back our original input, x.
Let's check f(f⁻¹(x)):
f(f⁻¹(x)) = f(x² + 5) = √((x² + 5) - 5) = √(x²) = |x|.
Notice we get |x|, not x! This is a critical point. Remember the domain restriction of our original function f(x) = √(x-5)? It was x ≥ 5. This restriction becomes the range restriction for the inverse function. The range of the original function, f(x), was y ≥ 0, which now becomes the domain restriction for our inverse function, f⁻¹(x). Therefore, x ≥ 0 for f⁻¹(x).
Since the domain of f⁻¹(x) is restricted to x ≥ 0, we can say that √(x²) = x because x is non-negative. Therefore, under this restriction, f(f⁻¹(x)) = x. Now, let's check f⁻¹(f(x)):
f⁻¹(f(x)) = f⁻¹(√(x-5)) = (√(x-5))² + 5 = (x - 5) + 5 = x.
This one works out perfectly! So, considering the domain restriction, our inverse function f⁻¹(x) = x² + 5 is indeed the correct inverse.
Addressing the Original Question
Now, let’s circle back to the multiple-choice options you presented. We found that the inverse function is f⁻¹(x) = x² + 5. However, none of the provided options exactly match this result. It looks like there might be a typo in the options provided, specifically with the exponents. The correct form of the inverse function involves x², not x³. It is also crucial to remember the domain restriction of x ≥ 0 for the inverse function to be fully accurate.
Let’s consider the option that's closest to our answer and discuss why the others are incorrect:
- A. f⁻¹(x) = -x³ + 5: This is incorrect because the power is 3 instead of 2, and there's a negative sign, which is not present in our solution.
- B. No inverse: This is incorrect because we found an inverse function, f⁻¹(x) = x² + 5, with the domain restriction x ≥ 0.
- C. f⁻¹(x) = x³ - 5: This is incorrect because the power is 3 instead of 2, and there's a subtraction instead of addition.
- D. f⁻¹(x) = x³ + 5: This is incorrect because the power is 3 instead of 2.
- E. f⁻¹(x) = x³ + 25: This is incorrect because the power is 3 instead of 2, and the constant term is wrong.
- F. None of the above: Given our derived answer, this appears to be the correct choice due to the typo in the exponents of the provided options.
Therefore, the most accurate answer, given the options, is F. None of the above, because our calculated inverse function f⁻¹(x) = x² + 5 doesn't match any of the given choices.
Key Takeaways
Finding the inverse of a function might seem tricky at first, but it becomes much easier with practice. Here are some key takeaways to keep in mind:
- Understand the definition of an inverse function: It