If f and f^-1 are both differentiable for all x, with f(3) = 5 and f'(3) = 7 , then write the equation of a line tangent to the graph of f^-1

Question

abbles · Answer

I literally have no clue..

zepdrix · Answer

f(3)=5, which means f^-1(5)=3, ya?

zepdrix · Answer

$$\large\rm y=mx+b$$We're trying to create some linear function which is tangent to the inverse function at ... oh you didn't specify a location. I assume it's at x=3, yes?

You can find the `slope` by using your formula for derivative inverse,$$\large\rm m=(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}$$

zepdrix · Answer

Oh maybe this tangent line is being created at x=5.
That would make more sense.

Fill in the details when you come back :D

abbles · Answer

Would the tan line be 1/7 for the slope?

abbles · Answer

Would the coordinate be (5, 3)

abbles · Answer

Would the equation be y - 3 = 1/7(x-5)

zepdrix · Answer

Ooo yes good job!$$\large\rm (f^{-1})'(5)=\frac{1}{f'(f^{-1}(5))}=\frac{1}{f'(3)}=\frac{1}{7}=m$$And since $\large\rm f^{-1}(5)=3$ we have the coordinate (5,3) for the inverse function.