Find the points on the graph y=x^3/2 - x^1/2 at which the tangent line is parallel to y-x=3.
The Attempt at a Solution
First I found that the derivative of y=x^3/2 - x^1/2 is 1x.
I then rewrote the other line as y = 3+x and found the derivative to be 1.
Then I set 1x = 1, therefore x = 1
Now I plugged it back into the original function and got y = 1^3/2-1^1/2 = 1
So the point is (1,1).
But I have no idea if I am anywhere near correct. Any help would be greatly appreciated.