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# Let $f(x) = \sqrt{1 - x^2}$ , $0 \le x \le 1$.(a) Find $f^{-1}$. How is it related to $f$ ?(b) Identify the graph of $f$ and explain your answer to part (a).

## (a) $y=f(x)=\sqrt{1-x^{2}} \quad(0 \leq x \leq 1 \text { and note that } y \geq 0) \Rightarrow$ $y^{2}=1-x^{2} \Rightarrow x^{2}=1-y^{2} \Rightarrow x=\sqrt{1-y^{2}}$ . So $f^{-1}(x)=\sqrt{1-x^{2}}, 0 \leq x \leq 1 .$ We see that $f^{-1}$ and $f$ are the same function.(b) The graph of $f$ is the portion of the circle $x^{2}+y^{2}=1$ with $0 \leq x \leq 1$ and $0 \leq y \leq 1$ (quarter-circle in the first quadrant). The graph of $f$ is symmetric with respect to the line $y=x,$ so its reflection about $y=x$ is itself, that is, $f^{-1}=f$

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here we have the function f of X, and let's start by finding F members. So I'm going to write f of X as why. And then I'm going to switch X and y so we get X equals the square root of one minus y squared. And then I'm going to isolate why. So we'll square both sides and then we'll subtract one from both sides. That gives us the opposite of why still on the right side. And then we could multiply both sides by negative one. So we have y squared equals one minus X squared, and then we could square root both sides. Typically, when we square root, we put plus or minus in front of the square root sign. But let's think about this. We were told that the original function has our domain between zero and one. That means the inverse has a range between zero and one, so wise, only positive. So we're going to eliminate the negative. So why I equals the square root of one minus X squared, and we can call that F in verse now. How does that relate to F F was the square root of one minus X squared F inverse is the square root of one minus X squared. They are the same. So why is that happening? Well, let's take a look at the graph. So did you know that F of X is actually the top half of a circle? If you ignore the domain part of it for a moment, if you ignore this part of it, you have the top half of a circle with radius one and center 00 So that would be this right here. However we're not, including the entire half circle were only including the part that goes from X equals zero toe one. So we only want this part right here. Thats f of X. Now, imagine taking that graph and reflecting it across the line. Y equals X to get its inverse. Isn't it going to look exactly the same

Oregon State University

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