- #1
KingNothing
- 882
- 4
I'm stuck on this problem. I am hoping someone can walk me through it or get me past my choking point. The problem states:
Two equations have two shared tangent lines between them. Find the equations of these tangent lines analytically.
[tex]g(x)=x^2[/tex]
[tex]f(x)=-x^2+6x-5[/tex]
The first step I took was to define what is needed for two curves to share a single tangent line:
-[tex]g'(x_0)=f'(x_1)[/tex] slopes at two different x-values must be the same
-the tangent line to [tex](x_0,y_0)[/tex] must pass through the point [tex](x_1,y_1)[/tex] on the other curve with the same slope
I found the general equation for the tangent of a line [tex]f(x)[/tex] at [tex]x[/tex] to be [tex]f'(x)k-xf'(x)+f(x)[/tex] where [tex]k[/tex] is to be substituted for [tex]x[/tex] in the final equation.
I did this with both of the functions in the problem, and set them equal. My final result for this was [tex]4xk-2x^2=6k-5[/tex] Assuming you leave k in there and just try to get the sides looking equal, I got nowhere. After substituting x for k, there is no solution.
Two equations have two shared tangent lines between them. Find the equations of these tangent lines analytically.
[tex]g(x)=x^2[/tex]
[tex]f(x)=-x^2+6x-5[/tex]
The first step I took was to define what is needed for two curves to share a single tangent line:
-[tex]g'(x_0)=f'(x_1)[/tex] slopes at two different x-values must be the same
-the tangent line to [tex](x_0,y_0)[/tex] must pass through the point [tex](x_1,y_1)[/tex] on the other curve with the same slope
I found the general equation for the tangent of a line [tex]f(x)[/tex] at [tex]x[/tex] to be [tex]f'(x)k-xf'(x)+f(x)[/tex] where [tex]k[/tex] is to be substituted for [tex]x[/tex] in the final equation.
I did this with both of the functions in the problem, and set them equal. My final result for this was [tex]4xk-2x^2=6k-5[/tex] Assuming you leave k in there and just try to get the sides looking equal, I got nowhere. After substituting x for k, there is no solution.
Last edited: