Inverse Of Function F(x) = X² - 6x + 5, X ≥ 3

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Alright guys, let's dive into finding the inverse of the function f(x) = x² - 6x + 5, where x ≥ 3. This is a classic math problem that combines completing the square, understanding domain restrictions, and the concept of inverse functions. Buckle up, because we're about to break it down step-by-step!

Understanding the Function and Its Domain

First, let’s get a good grip on what we're dealing with. We have a quadratic function f(x) = x² - 6x + 5. Now, quadratics are parabolas, and parabolas are notorious for not being one-to-one functions over their entire domain. Why does this matter? Well, to have an inverse, a function must be one-to-one, meaning it passes both the vertical and horizontal line tests. Our restriction, x ≥ 3, is super important because it ensures that we're only looking at half of the parabola, specifically the part where the function is increasing. This makes it one-to-one, and therefore, invertible. Think about it visually: if we didn't have this restriction, a horizontal line could intersect the parabola at two points, meaning there would be two different x values mapping to the same y value, violating the one-to-one property. This restriction essentially chops off the left side of the parabola, leaving us with a function that gracefully increases as x increases. This is a crucial point that often gets overlooked, but it's the key to why this inverse even exists in the first place. So, remember, the domain restriction isn't just some random detail – it's what makes the inverse possible! Without it, we'd be trying to find the inverse of something that doesn't have one, which would be like trying to fit a square peg into a round hole. Let's keep this in mind as we proceed through the steps of finding the inverse function. Understanding the initial constraints is paramount for arriving at the correct and meaningful solution. Now that we've laid this groundwork, let's move on to manipulating the function to get it into a more manageable form for finding its inverse.

Completing the Square

The next step in finding the inverse involves a bit of algebraic trickery: completing the square. Completing the square is a technique that allows us to rewrite a quadratic expression in a form that reveals its vertex. Remember, the vertex is the turning point of the parabola, and knowing its coordinates can be super helpful. So, let's take our function, f(x) = x² - 6x + 5, and complete the square. Here's how it goes:

  1. Focus on the x² and x terms: x² - 6x. We want to turn this into a perfect square trinomial. A perfect square trinomial is something that can be factored into (x + a)² or (x - a)².
  2. Take half of the coefficient of the x term, square it, and add and subtract it: The coefficient of our x term is -6. Half of -6 is -3, and (-3)² is 9. So, we add and subtract 9: f(x) = x² - 6x + 9 - 9 + 5
  3. Rewrite as a perfect square: Now we can rewrite the first three terms as a perfect square: f(x) = (x - 3)² - 9 + 5
  4. Simplify: Combine the constants: f(x) = (x - 3)² - 4

Alright! We've successfully completed the square. Our function is now in the form f(x) = (x - 3)² - 4. This form is incredibly useful because it tells us that the vertex of the parabola is at the point (3, -4). Remember that vertex? It's crucial because our domain restriction, x ≥ 3, means we're only looking at the part of the parabola to the right of the vertex. This confirms that our function is indeed one-to-one on this domain, and therefore, has an inverse. Completing the square isn't just some random algebraic manipulation; it gives us valuable insight into the behavior of the function, which is essential for understanding its inverse. With our function now in vertex form, we're ready to take the next step: swapping x and y.

Swapping x and y

Okay, the core concept of finding an inverse function is to swap the roles of x and y. In other words, wherever we see an x, we replace it with a y, and wherever we see a y (which is really f(x)), we replace it with an x. So, let's take our completed square form, f(x) = (x - 3)² - 4, and make the swap. Remember, f(x) is just another way of writing y, so we have:

y = (x - 3)² - 4

Now, swap x and y:

x = (y - 3)² - 4

That's it! We've successfully swapped x and y. This might seem like a small step, but it's a fundamental one. By swapping the variables, we're essentially reversing the operation of the original function. Think of it like this: if the original function takes an x value and spits out a y value, the inverse function takes that y value and spits out the original x value. Swapping x and y is the mathematical way of expressing this reversal. Now, our goal is to solve for y to express the inverse function in the familiar form y = something. This is where the algebra gets a little more involved, but don't worry, we'll take it step-by-step. Remember to keep the domain restriction in mind as we proceed because it will affect the sign we choose when we take the square root. With the variables swapped, we're now ready to isolate y and find the explicit form of the inverse function. Let's move on to solving for y.

Solving for y

The next challenge is to isolate y. We need to get y by itself on one side of the equation. So, let's take our equation from the previous step:

x = (y - 3)² - 4

  1. Add 4 to both sides: This will get rid of the -4 on the right side: x + 4 = (y - 3)²
  2. Take the square root of both sides: Remember that when we take the square root, we usually have to consider both the positive and negative roots. However, this is where our domain restriction, x ≥ 3, comes into play. Since x ≥ 3 in the original function, (x - 3) will always be non-negative. This means that when we find the inverse and solve for y, we only need to consider the positive square root. This is a crucial point that often gets missed! So: √(x + 4) = y - 3
  3. Add 3 to both sides: This will isolate y: y = √(x + 4) + 3

We've done it! We've successfully solved for y. This means we've found the inverse function. Remember to consider only the positive root due to the original domain restriction.

The Inverse Function

Alright, after all that work, we've arrived at the inverse function. We typically denote the inverse of f(x) as f⁻¹(x). So, based on our calculations, we have:

f⁻¹(x) = √(x + 4) + 3

But hold on, we're not quite done yet! It's crucial to consider the domain of the inverse function. The domain of the inverse function is the range of the original function. Since f(x) = (x - 3)² - 4 and x ≥ 3, the range of f(x) is y ≥ -4. Therefore, the domain of f⁻¹(x) is x ≥ -4. This makes sense because we can't take the square root of a negative number. So, to be completely precise, we should write the inverse function as:

f⁻¹(x) = √(x + 4) + 3, x ≥ -4

And that's it! We've successfully found the inverse function, including its domain. Remember, finding the inverse involves several steps: understanding the original function and its domain, completing the square (if necessary), swapping x and y, solving for y, and determining the domain of the inverse function. It's a process that requires careful attention to detail, but with practice, you'll become a pro at finding inverses. Great job, guys! You nailed it!

Summary of Steps

Let's recap the steps we took to find the inverse of f(x) = x² - 6x + 5, where x ≥ 3:

  1. Understand the function and its domain: We recognized that the domain restriction x ≥ 3 was crucial for the function to be one-to-one and invertible.
  2. Complete the square: We rewrote the function as f(x) = (x - 3)² - 4 to reveal its vertex.
  3. Swap x and y: We replaced f(x) with y and then swapped x and y to get x = (y - 3)² - 4.
  4. Solve for y: We isolated y to find y = √(x + 4) + 3.
  5. Determine the domain of the inverse: We recognized that the domain of the inverse function is x ≥ -4, based on the range of the original function.
  6. State the inverse function: We wrote the inverse function as f⁻¹(x) = √(x + 4) + 3, x ≥ -4.

By following these steps, you can find the inverse of many functions. Remember to always pay attention to the domain restrictions, as they play a critical role in the existence and form of the inverse function.