Understanding F(x)=(x-2)^2: Domain, Range, And Inverse

Hey guys! Today we're diving deep into a super common but important math concept: analyzing a quadratic function, specifically $f(x)=(x-2)^2$. We'll figure out its domain, range, and then get into the nitty-gritty of finding its inverse. This stuff is fundamental, and once you get the hang of it, you'll see it pop up everywhere, from calculus to physics problems. So, let's get this party started and break down $f(x)=(x-2)^2$ piece by piece!

Deconstructing $f(x)=(x-2)^2$: Domain and Range Exploration

Alright, let's kick things off by looking at the domain and range of our function, $f(x)=(x-2)^2$. First up, the domain. This is basically all the possible x-values that you can plug into the function and get a valid output. For $f(x)=(x-2)^2$, which is a polynomial, there are no weird restrictions like division by zero or square roots of negative numbers. You can literally plug in any real number for 'x' and the function will work perfectly. So, the domain of $f(x)=(x-2)^2$ is all real numbers. We can write this as $(-\infty, \infty)$ or $\mathbb{R}$. Now, let's talk about the range. The range is all the possible y-values (or $f(x)$ values) that the function can produce. Remember, $f(x)=(x-2)^2$ is a squared term. And when you square any real number, the result is always non-negative – it's either zero or positive. The minimum value this function can ever reach is 0, and that happens when $(x-2)=0$, which means $x=2$. Since the graph of $f(x)=(x-2)^2$ is a parabola that opens upwards with its vertex at $(2,0)$, the y-values go up from 0 infinitely. So, the range of $f(x)=(x-2)^2$ is all non-negative real numbers. We can write this as $[0, \infty)$. Understanding the domain and range is crucial because it sets the stage for everything else we're going to do, especially when we tackle the inverse function. It gives us the boundaries and the overall behavior of our function.

Finding the Sweet Spot: One-to-One and Non-Decreasing Domain

Now, here's where things get a bit more specific, guys. The question asks for the largest domain on which $f(x)=(x-2)^2$ is one-to-one and non-decreasing. Let's unpack that. A function is one-to-one if each output (y-value) corresponds to exactly one input (x-value). Think of it as a strict 'one-to-one' relationship – no two different x's can give you the same y. Our function $f(x)=(x-2)^2$ is a parabola, and parabolas are famously not one-to-one over their entire domain because, for any y-value greater than 0, there are two x-values that produce it (one on each side of the axis of symmetry). For example, if $f(x)=4$, then $(x-2)^2=4$, meaning $x-2 = \pm 2$, so $x=4$ or $x=0$. Both $x=0$ and $x=4$ give us $f(x)=4$. So, $f(x)$ isn't one-to-one globally.

Next, we need the function to be non-decreasing. A function is non-decreasing if, as you move from left to right along the x-axis, the y-values either stay the same or go up. They never go down. Looking at the graph of $f(x)=(x-2)^2$, which is a parabola with its vertex at $x=2$, the function decreases for $x < 2$ and increases for $x > 2$. Therefore, the interval where $f(x)$ is non-decreasing is $x \ge 2$. This is the part of the parabola that is going upwards.

So, we need a domain that satisfies both conditions: one-to-one and non-decreasing. The interval $x \ge 2$ fits the bill perfectly. On this interval, from $x=2$ onwards, the function's values are strictly increasing (which is a type of non-decreasing), and because we're only on one side of the vertex, each x-value gives a unique y-value. Thus, the largest domain on which $f(x)=(x-2)^2$ is one-to-one and non-decreasing is $[2, \infty)$. This is a super important step because when we find the inverse, we can only do it for a part of the function where it's behaving nicely like this.

Unveiling the Range of $f$ on the Restricted Domain

Now that we've pinned down the specific domain we're interested in – that sweet spot where $f(x)=(x-2)^2$ is both one-to-one and non-decreasing, which we found to be $[2, \infty)$ – let's figure out the range of $f$ on this particular domain. Remember, the range consists of all the possible output values (y-values) that the function produces when we only consider x-values from this restricted domain. We already established that the overall range of $f(x)=(x-2)^2$ is $[0, \infty)$ because it's a parabola opening upwards with its minimum at $y=0$.

Let's think about what happens when our input 'x' starts at 2 and goes towards positive infinity. When $x=2$, $f(2) = (2-2)^2 = 0^2 = 0$. So, the smallest output value we get from this domain is 0. As 'x' increases beyond 2 – think $x=3$, $x=4$, $x=100$, and so on – the term $(x-2)$ gets larger and larger, and squaring it makes it even larger. For instance, $f(3)=(3-2)^2 = 1^2 = 1$, $f(4)=(4-2)^2 = 2^2 = 4$, and if $x$ is a very large number, $(x-2)^2$ will also be a very large positive number.

Since we're on the domain $[2, \infty)$, where the function is non-decreasing (in fact, strictly increasing), the output values will start at $f(2)=0$ and increase without bound as 'x' heads towards infinity. Therefore, the range of $f$ restricted to the domain $[2, \infty)$ is precisely the set of all non-negative real numbers. We can express this range as $[0, \infty)$. It's important to note that this is the same as the overall range of the original function, but understanding how it arises from the restricted domain is key for finding the inverse.

The Art of Finding the Inverse: $f^{-1}(x)$

Okay, guys, this is where the magic happens – we're going to find the inverse function, $f^{-1}(x)$, for our restricted $f(x)=(x-2)^2$. Remember, we're only working with the part of the function where $x \ge 2$. The general process for finding an inverse function is to swap 'x' and 'y' and then solve for 'y'. Let's do that!

1. Start with the function: We're considering $y = (x-2)^2$ for $x \ge 2$.
2. Swap x and y: This gives us $x = (y-2)^2$.
3. Solve for y: Our goal now is to isolate 'y'. To get rid of the square, we'll take the square root of both sides. Remember, when you take the square root, you usually get a $\pm$ sign. So, $\sqrt{x} = \pm (y-2)$.
4. Isolate y: Now, rearrange the equation to get 'y' by itself. We have $\pm \sqrt{x} = y-2$. Adding 2 to both sides gives us $y = 2 \pm \sqrt{x}$.

Now, here's the critical part: we need to choose the correct sign ($\pm$). We know that the inverse function $f^{-1}(x)$ must have a domain and range that correspond to the range and domain of the restricted $f(x)$.

The restricted domain of $f$ was $[2, \infty)$, and its range was $[0, \infty)$.

Therefore, the domain of $f^{-1}$ must be $[0, \infty)$, and its range must be $[2, \infty)$.

Looking at our potential inverse $y = 2 \pm \sqrt{x}$, we need to select the option that gives us a range of $[2, \infty)$.

• If we choose the '+' sign: $y = 2 + \sqrt{x}$. Since $\sqrt{x}$ is always non-negative (and our domain for $f^{-1}$ is $x \ge 0$), $y$ will always be greater than or equal to 2. This matches our required range!
• If we choose the '-' sign: $y = 2 - \sqrt{x}$. This would give values less than or equal to 2, which doesn't match our required range of $[2, \infty)$.

So, the correct inverse function is $f^{-1}(x) = 2 + \sqrt{x}$.

We can check this: Pick an x from the restricted domain of $f$, say $x=3$. $f(3)=(3-2)^2 = 1$. Now plug this output (1) into our inverse: $f^{-1}(1) = 2 + \sqrt{1} = 2 + 1 = 3$. It works! We got our original input back.

Defining the Domain and Range of the Inverse Function

Finally, let's nail down the domain and range of our inverse function, $f^{-1}(x) = 2 + \sqrt{x}$. This is where understanding the relationship between a function and its inverse really shines, guys. The domain of the inverse function ($f^{-1}$) is always equal to the range of the original (restricted) function ($f$). And, conversely, the range of the inverse function ($f^{-1}$) is always equal to the domain of the original (restricted) function ($f$).

We already figured this out during the process of finding the inverse, but let's state it clearly:

• Domain of $f^{-1}$: We determined that the range of our restricted $f(x)=(x-2)^2$ (on the domain $[2, \infty)$) was $[0, \infty)$. Therefore, the domain of $f^{-1}(x) = 2 + \sqrt{x}$ is $[0, \infty)$. This makes sense intuitively too, because you can't take the square root of a negative number, and $\sqrt{x}$ requires $x \ge 0$.

• Range of $f^{-1}$: We determined that the domain of our restricted $f(x)=(x-2)^2$ was $[2, \infty)$. Therefore, the range of $f^{-1}(x) = 2 + \sqrt{x}$ is $[2, \infty)$. Again, this fits perfectly with our derived inverse function. Since $\sqrt{x}$ (for $x \ge 0$) produces values from $0$ upwards, adding 2 to it ($2 + \sqrt{x}$) means the output values will start at $2+0=2$ and go upwards towards infinity.

So, to recap: for $f(x)=(x-2)^2$ on the domain $[2, \infty)$, we have $f^{-1}(x) = 2 + \sqrt{x}$, with a domain of $[0, \infty)$ and a range of $[2, \infty)$. See how it all ties together? Mastering these concepts is super rewarding and really builds a solid foundation for more advanced math. Keep practicing, and you'll be an inverse function pro in no time!