Could some kind soul look this over to see if I made any mistakes?
Appreciate your comments on my problem areas (highlighted). thanks!
A function is defined by
Write down, where possible, the formula for the local inverse function at the points (x,y)=(1,0) and (-1,0), justifying your answers.
For the local inverse function at one of the points above sketch the region in the (u,v)-plane on which it is defined and its image on (x,y)-plane. Find the derivative of the local inverse function at the point you have chosen previously.
First, I'll find the critical points where the local inverse does not exist: det(J)=0
|2 -1| is the Jacobian
It's determinant det(J)=-2x-2=-2(x+1)=0 iff x=-1 which is a set of critical points, actually, a line: it divides the (x,y)-plane into two parts: J>0 to the left of the line x=-1, J<0 to the right of it.
(does the Jacobian's sign have any meaning or use in the context of finding a local inverse?)
So, of the two given points (1,0) and (-1,0), the local inverse only exists at (x,y)=(1,0).
To find local inverse at this point, I will solve the equations
for x and y.
I want x>0, therefore,
Therefore, the local inverse of f at (1,0) is described by the above equations and the point on the inverse correspoding to (x,y)=(1,0) is (u,v)=(1,2) (by using the definition of f).
Next, the region on (u,v) where the local inverse is defined is limited by
(since x>0) and I would draw a 45-degree line u=-v across II and IV quadrant of the (u,v)-plane, and the relevant region is the area above that line.
I am not sure the above region is correct.
For sketching coordinates of the inverse on (x,y)-plane, these are:y=1-x^2 (downward parabola crossing Y axis at y=1) and line y=2x-2. These two touch at a point (-1,0) - why is that? something to do with critical point or just a coincidence?
Finally, the derivative of the inverse is the inverted Jacobian matrix calculated at the point (x,y)=(1,0) which I got to
Thank you for your time!