Question

find the inverse of the following function. also list the domain and range restriction for the inverse.
$f(x) = \sqrt{-x^2 + 4}$, for $-2 \leq x \leq 0$
$f^{-1}(x) = \boxed{\quad}$
for domain $\boxed{\quad} \quad \boxed{\quad} \quad \boxed{\quad}$
and range $\boxed{\quad} \quad \boxed{\quad} \quad \boxed{\quad}$
options: $x^2 + 2$, $x^2 - 2$, $-x^2 + 2$, $-x^2 - 2$, $\sqrt{-x^2 + 4}$, $\sqrt{-x^2 - 4}$, $<$, $>$, $\leq$, $\geq$, $-4$, $-2$, $0$, $2$, $4$

Explanation:

Step1: Let \( y = f(x)=\sqrt{-x^{2}+4} \) for \( -2\leq x\leq0 \)

We start by replacing \( f(x) \) with \( y \). So we have \( y=\sqrt{-x^{2}+4} \), and the domain of the original function is \( -2\leq x\leq0 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap \( x \) and \( y \). So we get \( x = \sqrt{-y^{2}+4} \)

Step3: Solve for \( y \)

First, square both sides to get rid of the square root: \( x^{2}=-y^{2}+4 \)
Then, rearrange the equation to solve for \( y^{2} \): \( y^{2}=4 - x^{2} \)
Take the square root of both sides. But we need to consider the domain of the original function (\( -2\leq x\leq0 \)) to determine the sign of \( y \) in the inverse function. The original function \( f(x)=\sqrt{-x^{2}+4} \) for \( -2\leq x\leq0 \) has a range. Let's find the range of \( f(x) \): when \( x = - 2 \), \( y=\sqrt{-(-2)^{2}+4}=\sqrt{0} = 0 \); when \( x = 0 \), \( y=\sqrt{-0^{2}+4}=2 \). So the range of \( f(x) \) is \( 0\leq y\leq2 \). When we find the inverse, the domain of the inverse function is the range of the original function (\( 0\leq x\leq2 \)) and the range of the inverse function is the domain of the original function (\( - 2\leq y\leq0 \)). So when we solve \( y^{2}=4 - x^{2} \), we take the negative square root (because the original \( x \) values are non - positive, so the inverse function should give non - positive \( y \) values) \( y=-\sqrt{4 - x^{2}}=-\sqrt{-x^{2}+4} \)? Wait, no, let's re - do the solving.

Wait, from \( x^{2}=-y^{2}+4 \), we can rewrite it as \( y^{2}=4 - x^{2}\), so \( y=\pm\sqrt{4 - x^{2}} \). But the original function has \( x\in[-2,0] \), so when we find the inverse, the range of the inverse function is the domain of the original function, i.e., \( y\in[-2,0] \). So we take the negative square root? Wait, no, let's check with the original function. Let's take \( x=-2 \), \( y = 0 \); \( x = 0 \), \( y = 2 \). So the original function is increasing on \( [-2,0] \) (we can check the derivative or test points: for \( x_1=-2,x_2 = - 1 \), \( f(-2)=0,f(-1)=\sqrt{-1 + 4}=\sqrt{3}\approx1.732 \), so as \( x \) increases from - 2 to 0, \( y \) increases from 0 to 2). So the inverse function should be a function where when \( x \) is in the range of \( f(x) \) (i.e., \( 0\leq x\leq2 \)), \( y \) is in \( [-2,0] \) and is a decreasing function? Wait, maybe I made a mistake in the sign. Let's start over.

Original function: \( y=\sqrt{-x^{2}+4} \), \( -2\leq x\leq0 \)

Square both sides: \( y^{2}=-x^{2}+4\Rightarrow x^{2}=4 - y^{2}\Rightarrow x = \pm\sqrt{4 - y^{2}} \)

But since \( x\in[-2,0] \), \( x \) is non - positive, so \( x=-\sqrt{4 - y^{2}} \)

Now swap \( x \) and \( y \) to get the inverse function: \( y =-\sqrt{4 - x^{2}}=-\sqrt{-x^{2}+4} \)? Wait, no, \( 4 - y^{2}=-x^{2}+4 \), so \( x^{2}=4 - y^{2}\), so \( x =-\sqrt{4 - y^{2}} \) (because \( x\leq0 \)). Then the inverse function \( f^{-1}(x)=-\sqrt{4 - x^{2}} \)? Wait, but let's check the domain and range.

The domain of \( f(x) \) is \( [-2,0] \), so the range of \( f^{-1}(x) \) is \( [-2,0] \)

The range of \( f(x) \): when \( x=-2 \), \( y = 0 \); when \( x = 0 \), \( y = 2 \), so the range of \( f(x) \) is \( [0,2] \), so the domain of \( f^{-1}(x) \) is \( [0,2] \)

Now, let's simplify \( f^{-1}(x) \). We have \( y =-\sqrt{4 - x^{2}} \), but we can also rewrite \( 4 - x^{2}=-x^{2}+4 \), so \( f^{-1}(x)=-\sqrt{-x^{2}+4} \)? Wait, no, let's check the options. Wait, the options have \( -x^{2}+2 \)? No, maybe I made a mistake. Wait, the original function is \( y = \sqrt{-x^{2}+4}\), let's consider the graph. The original function is the upper half of the circle \( x^{2}+y^{2}=4 \) (since \( y\geq0 \)) with \( x\in[-2,0] \), so it's the left - hand semicircle of the upper half - circle. The inverse function should be the lower half of the circle \( x^{2}+y^{2}=4 \) with \( x\in[0,2] \), so \( y=-\sqrt{4 - x^{2}} \), but let's check the options. Wait, the options have \( -x^{2}+2 \)? No, maybe the original function is misread. Wait, the original function is \( f(x)=\sqrt{-x^{2}+4} \), maybe it's a semicircle. Wait, another approach: let's find the domain and range of the inverse function.

Domain of \( f^{-1}(x) \): range of \( f(x) \). For \( f(x)=\sqrt{-x^{2}+4} \), \( -2\leq x\leq0 \). When \( x=-2 \), \( f(-2)=0 \); when \( x = 0 \), \( f(0)=2 \). And since \( y=\sqrt{-x^{2}+4} \) is increasing on \( [-2,0] \) (as \( x \) increases, \( -x^{2}+4 \) increases, so the square root increases), the range of \( f(x) \) is \( 0\leq y\leq2 \). So the domain of \( f^{-1}(x) \) is \( 0\leq x\leq2 \).

Range of \( f^{-1}(x) \): domain of \( f(x) \), which is \( -2\leq y\leq0 \).

Now, let's find the inverse function. Let \( y = f(x)=\sqrt{-x^{2}+4} \), \( -2\leq x\leq0 \)

1. Swap \( x \) and \( y \): \( x=\sqrt{-y^{2}+4} \), \( 0\leq y\leq2 \) (since \( y \) is in the range of \( f(x) \))
2. Square both sides: \( x^{2}=-y^{2}+4 \)
3. Solve for \( y \): \( y^{2}=4 - x^{2}\Rightarrow y=\pm\sqrt{4 - x^{2}} \)
4. Since \( y \) is in \( [-2,0] \) (range of \( f^{-1}(x) \)), we take the negative square root: \( y =-\sqrt{4 - x^{2}}=-\sqrt{-x^{2}+4} \)? Wait, no, \( 4 - x^{2}=-y^{2}+4 \), so \( y^{2}=4 - x^{2}\), so \( y =-\sqrt{4 - x^{2}} \) (because \( y\in[-2,0] \))

But let's check with \( x = 0 \): \( f(0)=2 \), so \( f^{-1}(2)=0 \). Let's plug \( x = 2 \) into \( y=-\sqrt{4 - x^{2}} \), we get \( y = 0 \); plug \( x = 0 \) into \( y=-\sqrt{4 - x^{2}} \), we get \( y=-2 \). Wait, that's not correct. Wait, I think I messed up the sign. Let's go back. The original function: \( x\in[-2,0] \), \( y\in[0,2] \). When we swap \( x \) and \( y \), we have \( y\in[-2,0] \), \( x\in[0,2] \). So when we have \( x=\sqrt{-y^{2}+4} \), \( y\in[-2,0] \), so \( -y^{2}+4\geq0\Rightarrow y^{2}\leq4\Rightarrow - 2\leq y\leq2 \), which is consistent. Now, solve for \( y \):

\( x^{2}=-y^{2}+4\Rightarrow y^{2}=4 - x^{2}\Rightarrow y = \pm\sqrt{4 - x^{2}} \)

But since \( y\in[-2,0] \), \( y=-\sqrt{4 - x^{2}} \) would give \( y\in[-2,0] \) when \( x\in[0,2] \) (because when \( x = 0 \), \( y=-2 \); when \( x = 2 \), \( y = 0 \)). But the original function at \( x=-2 \) gives \( y = 0 \), so \( f^{-1}(0)=-2 \). Let's check \( f^{-1}(0) \): if \( f^{-1}(0)=-2 \), then \( f(-2)=0 \), which is correct. If we take \( y =-\sqrt{4 - x^{2}} \), when \( x = 0 \), \( y=-2 \), which is \( f^{-1}(0)=-2 \), correct. When \( x = 2 \), \( y = 0 \), which is \( f^{-1}(2)=0 \), correct.

Now, let's check the options. The options for the inverse function: we have \( -x^{2}+2 \)? No, maybe the original function is a semicircle of a different kind. Wait, maybe the original function is \( y=\sqrt{-x^{2}+4} \), which is the upper half of the circle \( x^{2}+y^{2}=4 \), and the domain \( -2\leq x\leq0 \) is the left - hand upper semicircle. The inverse function should be the lower half of the circle \( x^{2}+y^{2}=4 \) with \( x\in[0,2] \), which is \( y =-\sqrt{4 - x^{2}} \), but let's rewrite \( -\sqrt{4 - x^{2}}=-\sqrt{-x^{2}+4} \). But looking at the options, we have \( -x^{2}+2 \)? No, maybe I made a mistake in the function. Wait, maybe the original function is \( y=\sqrt{-x^{2}+4} \), and when we solve for the inverse, we can also note that \( y^{2}=-x^{2}+4\Rightarrow x^{2}+y^{2}=4 \), which is a circle. The original function is the left - hand upper quarter - circle, and the inverse function is the lower - hand right - quarter circle.

Now, let's check the domain and range of the inverse function. Domain of \( f^{-1}(x) \): \( 0\leq x\leq2 \) (since it's the range of \( f(x) \)). Range of \( f^{-1}(x) \): \( -2\leq y\leq0 \) (since it's the domain of \( f(x) \)).

Now, let's find the inverse function again. Let \( y=\sqrt{-x^{2}+4} \), \( -2\leq x\leq0 \)

1. Swap \( x \) and \( y \): \( x=\sqrt{-y^{2}+4} \), \( 0\leq x\leq2 \)
2. Square both sides: \( x^{2}=-y^{2}+4 \)
3. Rearrange: \( y^{2}=4 - x^{2}\Rightarrow y = \pm\sqrt{4 - x^{2}} \)
4. Since \( y\in[-2,0] \), we take \( y=-\sqrt{4 - x^{2}} \)

Now, let's check the options. Wait, maybe the original function is \( y=\sqrt{-x^{2}+4} \), and we can also write \( -x^{2}+4=4 - x^{2} \), and the inverse function is \( y =-\sqrt{4 - x^{2}} \), but looking at the options, we have \( -x^{2}+2 \)? No, maybe the function is different. Wait, maybe the original function is \( y=\sqrt{-x^{2}+4} \), and when we solve for the inverse, we can also note that \( x^{2}=4 - y^{2}\Rightarrow y^{2}=4 - x^{2}\Rightarrow y =-\sqrt{4 - x^{2}} \), but let's check with \( x = 0 \), \( f(0)=2 \), so \( f^{-1}(2)=0 \). Plug \( x = 2 \) into \( y=-\sqrt{4 - x^{2}} \), we get \( y = 0 \); plug \( x = 0 \) into \( y=-\sqrt{4 - x^{2}} \), we get \( y=-2 \), which is correct because \( f(-2)=0 \), so \( f^{-1}(0)=-2 \).

Now, for the domain of \( f^{-1}(x) \): since the range of \( f(x) \) is \( 0\leq y\leq2 \), the domain of \( f^{-1}(x) \) is \( 0\leq x\leq2 \) (or \( 0\leq x\leq2 \), using the symbol \( \leq \)).

For the range of \( f^{-1}(x) \): since the domain of \( f(x) \) is \( -2\leq x\leq0 \), the range of \( f^{-1}(x) \) is \( -2\leq y\leq0 \).

Now, the inverse function \( f^{-1}(x)=-\sqrt{-x^{2}+4} \)? Wait, no, \( \sqrt{-x^{2}+4} \) is the original function. Wait, I think I made a mistake in the algebra. Let's do it again:

Start with \( y=\sqrt{-x^{2}+4} \), \( -2\leq x\leq0 \)

Square both sides: \( y^{2}=-x^{2}+4\Rightarrow x^{2}=4 - y^{2}\Rightarrow x = \pm\sqrt{4 - y^{2}} \)

Since \( x\in[-2,0] \), \( x=-\sqrt{4 - y^{2}} \)

Now, swap \( x \) and \( y \): \( y =-\sqrt{4 - x^{2}} \), \( 0\leq x\leq2 \)

So \( f^{-1}(x)=-\sqrt{4 - x^{2}}=-\sqrt{-x^{2}+4} \)

Now, let's check the domain of \( f^{-1}(x) \): \( 0\leq x\leq2 \) (because the range of \( f(x) \) is \( 0\leq y\leq2 \))

Range of \( f^{-1}(x) \): \( -2\leq y\leq0 \) (because the domain of \( f(x) \) is \( -2\leq x\leq0 \))

Answer:

The inverse function \( f^{-1}(x)=-\sqrt{-x^{2}+4} \)

Domain of \( f^{-1}(x) \): \( 0\leq x\leq2 \)

Range of \( f^{-1}(x) \): \( -2\leq y\leq0 \)