Finding the inverse function of
![\mathsf{f(x)=10-x^2}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%28x%29%3D10-x%5E2%7D)
Remember that when you compose
![f](https://tex.z-dn.net/?f=f)
with its inverse
![\mathsf{f^{-1}},](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5E%7B-1%7D%7D%2C)
you'll get the identity function:
![\mathsf{(f\circ f^{-1})(x)=x}\\\\ \mathsf{f\big[f^{-1}(x)\big]=x}\\\\](https://tex.z-dn.net/?f=%5Cmathsf%7B%28f%5Ccirc%20f%5E%7B-1%7D%29%28x%29%3Dx%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3Dx%7D%5C%5C%5C%5C)
So if
![\mathsf{f(x)=10-x^2}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%28x%29%3D10-x%5E2%7D)
then
![\mathsf{f\big[f^{-1}(x)\big]=10-[f^{-1}(x)]^2}\\\\ \mathsf{x=10-[f^{-1}(x)]^2}\\\\ \mathsf{[f^{-1}(x)]^2=10-x}\\\\ \mathsf{f^{-1}(x)=\pm \sqrt{10-x}}\\\\ \mathsf{f^{-1}(x)=-\sqrt{10-x}~~~or~~~f^{-1}(x)=\sqrt{10-x}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7Bx%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Bf%5E%7B-1%7D%28x%29%5D%5E2%3D10-x%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D%5Cpm%20%5Csqrt%7B10-x%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D-%5Csqrt%7B10-x%7D~~~or~~~f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B10-x%7D%7D)
The sign of the inverse depends on the domain of
![\mathsf{f(x)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%28x%29%7D)
itself, and where it's invertible.
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Tags: <em>inverse function definition identity domain algebra</em>
Answer:
Step-by-step explanation:
If you have trouble working with the coordinate numbers, it can be helpful to plot them on a graph so that you can count grid squares. (see attached)
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The translation is ...
A' -A = (2, 3) -(3, -4) = (2-3, 3-(-4)) = (-1, 7)
Since positive numbers are to the right or up, negative numbers are to the left or down.
The translation (-1, 7) means a translation 1 unit left and 7 units up.
Answer:
its many solutions
Step-by-step explanation: