Find The Inverse Of F(x)=sqrt(3x-2)

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a problem that might make some of you scratch your heads: finding the inverse of a function. We're going to break down the inverse of $f(x)=\sqrt{3x-2}$ for $x \geq \frac{2}{3}$. This isn't just about getting the right answer; it's about understanding the why and how behind it. So, grab your calculators, maybe a coffee, and let's get this done!

Understanding Inverse Functions

Before we jump into solving for the inverse of $f(x)=\sqrt{3x-2}$, let's make sure we're all on the same page about what an inverse function actually is. Think of a function like a machine. You put something in (the input, usually $x$), and it does something to it to give you something out (the output, usually $f(x)$ or $y$). An inverse function, denoted as $f^{-1}(x)$, is like a reverse machine. If you feed the output of the original function back into the inverse function, you get the original input back. Essentially, it undoes what the original function did. For a function to have an inverse, it must be one-to-one, meaning each output corresponds to only one unique input. In our case, $f(x)=\sqrt{3x-2}$ with the domain $x \geq \frac{2}{3}$ is indeed one-to-one because the square root function, when restricted to a domain where it's increasing, produces unique outputs for each unique input. The domain restriction $x \geq \frac{2}{3}$ is crucial because it ensures that the expression inside the square root, $3x-2$, is non-negative, which is a requirement for the square root of a real number. When $x=\frac{2}{3}$, $3x-2 = 3(\frac{2}{3})-2 = 2-2 = 0$, so $f(\frac{2}{3}) = \sqrt{0} = 0$. As $x$ increases from $\frac{2}{3}$, $3x-2$ increases, and so does its square root. This monotonic behavior (always increasing) is what guarantees the function is one-to-one over its specified domain.

Steps to Find the Inverse Function

Alright, let's get down to business. Finding the inverse function, $f^{-1}(x)$, involves a few key steps. We'll use $y$ to represent $f(x)$ initially, which is common practice. So, our function is $y = \sqrt{3x-2}$. The first step is to swap $x$ and $y$. This is the core idea of finding an inverse – we're essentially saying, 'what input ($x$) would give me this output ($y$)?' So, we rewrite the equation as $x = \sqrt{3y-2}$. Notice how we replaced $y$ with $x$ and $x$ with $y$. This swapped equation now represents the relationship for the inverse function, but we need to solve it for $y$ to get it in the standard $f^{-1}(x)$ form. The next step is to isolate $y$. This is where the algebra comes in. To get rid of the square root, we need to square both sides of the equation. So, $x^2 = (\sqrt{3y-2})^2$, which simplifies to $x^2 = 3y-2$. Now, we want to get $3y$ by itself, so we add 2 to both sides: $x^2 + 2 = 3y$. Finally, to get $y$ completely isolated, we divide both sides by 3: $y = \frac{x^2+2}{3}$. This is our potential inverse function. But wait, there's a crucial part we can't forget: the domain and range of the inverse function. The domain of the original function $f(x)$ becomes the range of the inverse function $f^{-1}(x)$, and the range of the original function $f(x)$ becomes the domain of the inverse function $f^{-1}(x)$. Let's figure out the range of $f(x)=\sqrt{3x-2}$ for $x \geq \frac{2}{3}$. As we established, when $x=\frac{2}{3}$, $f(x)=0$. As $x$ increases, $f(x)$ also increases without bound. So, the range of $f(x)$ is $[0, \infty)$, or $f(x) \geq 0$. Since the range of $f(x)$ is the domain of $f^{-1}(x)$, this means the domain of our inverse function is $x \geq 0$. Therefore, our inverse function is $f^{-1}(x) = \frac{1}{3}(x^2+2)$ with the domain $x \geq 0$. This matches option C, which is awesome!

Analyzing the Options

Let's quickly look at the options provided to solidify our understanding and see why option C is the correct one. We have:

A. $f^{-1}(x)=\frac{1}{3}\left(x^2+2\right), x \leq 0$

B. $f^1(r)=\frac{1}{2}\left(r^2+9\right) x$

C. $f^{-1}(x)=\frac{1}{3}\left(x^2+2\right), x \geq 0$

We've already derived that $f^{-1}(x) = \frac{1}{3}(x^2+2)$ and determined that its domain must be $x \geq 0$. Option A has the correct function form but the incorrect domain ($x \leq 0$). If we were to use this domain, it would mean we're only considering negative inputs for the inverse, which doesn't align with the range of our original function. Option B looks completely different. It uses $r$ instead of $x$ as the variable and has a $\frac{1}{2}$ instead of $\frac{1}{3}$, and even includes an extra $x$ term, which suggests it's likely an incorrect transformation or perhaps a typo. The coefficients and structure don't match our derivation at all. Option C, however, perfectly matches our derived inverse function $f^{-1}(x)=\frac{1}{3}(x^2+2)$ and includes the correct domain $x \geq 0$. This domain is derived from the range of the original function $f(x)$. Remember, the range of $f(x) = \sqrt{3x-2}$ for $x \geq \frac{2}{3}$ is $[0, \infty)$, meaning $f(x) \geq 0$. Since the domain of $f^{-1}(x)$ is the range of $f(x)$, the domain of $f^{-1}(x)$ must be $x \geq 0$. So, option C is our winner, guys!

Verifying the Inverse

To be absolutely sure, let's do a quick check by plugging values into both $f(x)$ and $f^{-1}(x)$ to see if we get the expected results. Remember, for a function and its inverse, $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. Let's pick a value for $x$ in the domain of $f(x)$, say $x=2$. Since $2 \geq \frac{2}{3}$, it's a valid input. First, find $f(2)$: $f(2) = \sqrt{3(2)-2} = \sqrt{6-2} = \sqrt{4} = 2$. Now, let's take this output, $2$, and plug it into our inverse function $f^{-1}(x) = \frac{1}{3}(x^2+2)$. Since $2 \geq 0$, it's a valid input for $f^{-1}(x)$. So, $f^{-1}(2) = \frac{1}{3}(2^2+2) = \frac{1}{3}(4+2) = \frac{1}{3}(6) = 2$. And voilà! We got our original input $x=2$ back. This confirms that our inverse function works for this specific value. Let's try another one, maybe $x=6$. $f(6) = \sqrt{3(6)-2} = \sqrt{18-2} = \sqrt{16} = 4$. Now, we plug $4$ into $f^{-1}(x)$. Since $4 \geq 0$, it's valid. $f^{-1}(4) = \frac{1}{3}(4^2+2) = \frac{1}{3}(16+2) = \frac{1}{3}(18) = 6$. Again, we get our original input back! This process of checking $f^{-1}(f(x)) = x$ helps build confidence in our answer. We could also check $f(f^{-1}(x)) = x$. Let's pick an $x$ from the domain of $f^{-1}(x)$, which is $x \geq 0$. Let's use $x=3$. $f^{-1}(3) = \frac{1}{3}(3^2+2) = \frac{1}{3}(9+2) = \frac{1}{3}(11) = \frac{11}{3}$. Now, we plug $\frac{11}{3}$ into $f(x)$. We need to check if $\frac{11}{3} \geq \frac{2}{3}$, which it is. $f(\frac{11}{3}) = \sqrt{3(\frac{11}{3})-2} = \sqrt{11-2} = \sqrt{9} = 3$. We got our original input $x=3$ back! These checks are super important in mathematics to ensure your calculations are correct. It’s like double-checking your work before submitting an assignment.

Conclusion

So there you have it, folks! Finding the inverse of a function like $f(x)=\sqrt{3x-2}$ involves a systematic approach: swapping $x$ and $y$, then solving for the new $y$. But don't forget the critical step of determining the correct domain for your inverse function, which is dictated by the range of the original function. In this case, the inverse of $f(x)=\sqrt{3x-2}$ for $x \geq \frac{2}{3}$ is indeed $f^{-1}(x)=\frac{1}{3}(x^2+2)$ with the domain $x \geq 0$. This corresponds to option C. Keep practicing these kinds of problems, and soon you'll be finding inverses like a pro! Math can be tricky, but breaking it down step-by-step makes it totally manageable. Stay tuned to Plastik Magazine for more math breakdowns and awesome content!