Question

Find the equation of the tangent to the graph of y+f^(-1) at x=1 where f(x)= x^(3)+x+1.

Can someone show me the steps to this please. Do I first find the inverse function and then get the derivative of that and then plug in x=1. I tried a couple of ways but I keep confusing myself

Answer 1

This might help you: http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation

Answer 2

This is the part that's confusing me. "The graph of: "
\[y + f ^{-1}\]
What is y? And how is that graphable? That's just an expression. What is the argument for f inverse? Is it f inverse of x? f inverse of y?

Answer 3

sorry that should say y=f^(-1) not +

Answer 4

Did the link help you?

Answer 5

sort of but I'm still a little messed up on my steps

Answer 6

Okay that makes more sense. Can I assume it's
\[f ^{-1}(x)\
?

Answer 7

yes

Answer 8

The steps you've listed make sense, but finding the explicit inverse function seems pretty tough, eh?

Answer 9

yea i seem to get mixed up in the numbers and im having a hard time finding the actual inverse

Answer 10

I don't think that finding the explicit inverse function is actually the solution in this case. It's simple to just look at the specific point give, x=1

Answer 11

and how do i use that point

Answer 12

Okay, so the function
\[f^{-1}(x)\]
takes in outputs of f(x) and puts out the corresponding input of f(x) that gives that output.
So:
\[f^{-1}(1)\]
Will be the x that gives f(x) = 1. By inspection, f(0) = 0^3 + 0 +1 = 1, so
\[f^{-1}(1) = 0\]

Answer 13

F(x) and f inverse look like: