Inverse Of Sqrt(4-x^2) For -2 <= X <= 0

Let's dive into the fascinating world of inverse functions with a specific example: $f(x)=\sqrt{4-x^2}$ defined over the interval $-2 \leq x \leq 0$. Finding the inverse of a function can seem like a daunting task at first, but with a systematic approach, it becomes quite manageable. The core idea behind an inverse function, often denoted as $f^{-1}(x)$, is that it "undoes" what the original function $f(x)$ does. If you apply a function and then its inverse, you should end up back where you started. Mathematically, this means that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ for all valid inputs in their respective domains.

Understanding the Original Function

Before we can find the inverse, it's crucial to understand the properties of our original function, $f(x)=\sqrt{4-x^2}$, over the given domain $-2 \leq x \leq 0$. This function describes the upper-left quarter of a circle centered at the origin with a radius of 2. Let's analyze this. Squaring both sides of $y = \sqrt{4-x^2}$, we get $y^2 = 4-x^2$, which rearranges to $x^2 + y^2 = 4$. This is the equation of a circle with radius 2. However, the original function specifies $y = \sqrt{4-x^2}$, meaning $y$ must be non-negative. Furthermore, the domain is restricted to $-2 \leq x \leq 0$. This restriction means we are only considering the part of the circle that lies in the second quadrant (where x is negative and y is positive) and includes the points $(-2, 0)$ and $(0, 2)$.

So, for our function $f(x)$, the input $x$ ranges from $-2$ to $0$. What about the output, $f(x)$? When $x = -2$, $f(-2) = \sqrt{4 - (-2)^2} = \sqrt{4-4} = 0$. When $x = 0$, $f(0) = \sqrt{4 - 0^2} = \sqrt{4} = 2$. Since the function represents the upper-left quarter circle, the output values $f(x)$ range from $0$ to $2$. Therefore, the domain of $f$ is $[-2, 0]$ and its range is $[0, 2]$.

Steps to Find the Inverse Function

Now, let's find the inverse function $f^{-1}(x)$. The standard procedure involves three main steps:

1. Replace $f(x)$ with $y$: So, $y = \sqrt{4-x^2}$.
2. Swap $x$ and $y$: This is the key step in finding an inverse. The equation becomes $x = \sqrt{4-y^2}$.
3. Solve for $y$: This will give us our inverse function, $f^{-1}(x)$.

Let's perform step 3. We have $x = \sqrt{4-y^2}$. To isolate $y$, we first square both sides:

$x^2 = 4-y^2$

Now, rearrange the equation to solve for $y^2$:

$y^2 = 4-x^2$

Taking the square root of both sides, we get:

$y = \pm\sqrt{4-x^2}$

Determining the Correct Sign and Domain

This is where we need to be careful. We have two possibilities for $y$: $y = \sqrt{4-x^2}$ and $y = -\sqrt{4-x^2}$. To determine which one is correct, we need to consider the domain and range of the original function and its inverse. Remember, the domain of the original function $f(x)$ becomes the range of its inverse $f^{-1}(x)$, and the range of $f(x)$ becomes the domain of $f^{-1}(x)$.

From our analysis earlier, the domain of $f$ is $[-2, 0]$ and the range of $f$ is $[0, 2]$. Therefore, the domain of $f^{-1}(x)$ must be $[0, 2]$, and the range of $f^{-1}(x)$ must be $[-2, 0]$.

Let's examine our two potential inverse functions:

• Case 1: $y = \sqrt{4-x^2}$ If we consider this as our inverse, its domain would be where $4-x^2 less 0$, so $-2 less x less 2$. The outputs $y$ would be non-negative. This doesn't match the required range of $[-2, 0]$ for our inverse function.

• Case 2: $y = -\sqrt{4-x^2}$ For this function, the expression under the square root, $4-x^2$, must be non-negative, so $-2 less x less 2$. The square root $\sqrt{4-x^2}$ will always yield a non-negative value. However, because of the negative sign in front, the output $y$ will always be non-positive (i.e., $y \leq 0$). This matches the required range of $[-2, 0]$ for our inverse function.

Now, let's consider the domain for this inverse. The domain of $f^{-1}(x)$ must be the range of $f(x)$, which is $[0, 2]$. So, we must have $0 \leq x \leq 2$. For these values of $x$, $4-x^2$ is indeed non-negative, and $-\sqrt{4-x^2}$ will produce values between $-2$ and $0$. For example, if $x=0$, $y = -\sqrt{4-0} = -2$. If $x=2$, $y = -\sqrt{4-4} = 0$. This aligns perfectly with the domain and range requirements.

Conclusion

Therefore, the inverse function $f^{-1}(x)$ for $f(x)=\sqrt{4-x^2}$ defined for $-2 \leq x \leq 0$ is $f^{-1}(x) = -\sqrt{4-x^2}$, and its domain is $0 \leq x \leq 2$. This makes sense because the original function maps the interval $[-2, 0]$ to $[0, 2]$, and the inverse function must map $[0, 2]$ back to $[-2, 0]$. The negative sign in front of the square root ensures that the output values are negative or zero, fitting within the required range.

To verify, let's pick a value from the domain of $f$, say $x=-1$. Then $f(-1) = \sqrt{4-(-1)^2} = \sqrt{4-1} = \sqrt{3}$. Now, let's apply the inverse function to $f(-1) = \sqrt{3}$. Remember, the input to the inverse function is the output of the original function, so we input $\sqrt{3}$ into $f^{-1}(x)$. Since $0 \leq \sqrt{3} \leq 2$, it is in the domain of $f^{-1}(x)$.

$f^{-1}(\sqrt{3}) = -\sqrt{4-(\sqrt{3})^2} = -\sqrt{4-3} = -\sqrt{1} = -1$. Indeed, $f^{-1}(f(-1)) = -1$, which is our original input $x$. This confirms our derived inverse function is correct.

If you're interested in learning more about functions and their inverses, exploring resources on calculus or algebra can provide a deeper understanding. Websites like Khan Academy offer excellent free courses and explanations on these topics.