Answer:
Point of tangency is [tex]x=1[/tex] or [tex](1,1)[/tex]
Step-by-step explanation:
You don't really need to use Calculus here. You can just set each equation to each other and solve for x (the point of tangency):
[tex]x^2=2x-1[/tex]
[tex]x^2-2x+1=0[/tex]
[tex](x-1)(x-1)=0[/tex]
[tex]x=1[/tex]
Therefore, the point of tangency is [tex]x=1[/tex] or [tex](1,1)[/tex]
