Having a bit of a problem getting the inverse of the following equation:
$$f(x) = \sqrt{9-x^2}$$
I had an answer which was equal to $3-x$ but when I used sites like Mathway and Wolfram to check my answer it said "No Inverse Equation".
Can anyone please tell me how this is so? Or is Mathway and Wolfram just mistaken?
On
With no further information, I will assume your domain is $[-3,3]$.
Then there is no inverse function because the function is not bijective.
On
The equation $\sqrt{9-x^2}$ represents half a circle. Inverse function means mirroring it by $y=x$ line. Doing so leads to $x=\sqrt{9-y^2}$ which is not a function. Because for example for $x=1$ it leads to two $y$, $y=\sqrt{8}$ and $-\sqrt{8}$.
On
Its true that this function doesn't have an inverse on its largest domain of definition $[-3,3]$ (Note that f(x)=f(-x)). But it does have on the two subintervals $[-3,0]$ and $[0,3]$. Lets calculate the latter. Here we have that: \begin{align*} y = \sqrt{9-x^2} \Leftrightarrow y^2 =9-x^2 \Leftrightarrow x^2 = 9-y^2 \Leftrightarrow x= \sqrt{9-y^2}. \end{align*} This isn't surprising given the geometric nature of the graph of $f$ (upper semicircle with center (0,0) and radius 3).
On
You start with:
$$y = \sqrt{9-x^2}$$
squaring both sides you get:
$$y^2 = {9-x^2}$$
then, multiply everything by $-1$:
$$-y^2 = {-9+x^2}$$
isolate x:
$$9-y^2 = {x^2}$$
and you get two possible inputs which might give the same result:
$$\pm\sqrt{9-y^2} = {x}$$
and the final line of code here is your inverse function.
If the function is defined as $f:[0,3]\to[0,3]$ where $f(x)=\sqrt{9-x^2}$ for all $x\in[0,3]$, then the inverse function is $f^{-1}:[0,3]\to[0,3]$ given by $f^{-1}(x)=\sqrt{9-x^2}$ for all $x\in[0,3]$.