# Thread: equation for the tangent

1. ## equation for the tangent

Hey guys, I completely forgot how to do this problem.

Find an equation for the line tangent to the graph of $f^{-1}$ at the point (3,1) if f(x)= $x^3$+2 $x^2$-x+1.

I'm not sure if a formal proof is needed and if so how to do it.

2. Originally Posted by alice8675309
Hey guys, I completely forgot how to do this problem, and its from the same book I've been using that's I'm still posting it in this section.

Find an equation for the line tangent to the graph of $f^{-1}$ at the point (3,1) if f(x)= $x^3$+2 $x^2$-x+1.

I'm not sure if a formal proof is needed and if so how to do it.
The tangent at the point $(3,1)$ of the graph $y=f^{-1}(x)$ is $t:\, y=f'^{-1}(3)\cdot (x-3)+1$.
The remaining problem is to determine $f'^{-1}(3)$. Luckily there is a formula that allows you to determine $f'^{-1}(x)$ from the derivatives of $f(x)$ like this: $f'^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$.
So in your case this means that $f'^{-1}(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\ldots$

3. Originally Posted by Failure
The tangent at the point $(3,1)$ of the graph $y=f^{-1}(x)$ is $t:\, y=f'^{-1}(3)\cdot (x-3)+1$.
The remaining problem is to determine $f'^{-1}(3)$. Luckily there is a formula that allows you to determine $f'^{-1}(x)$ from the derivatives of $f(x)$ like this: $f'^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$.
So in your case this means that $f'^{-1}(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\ldots$
Hey thanks so much, mental block there..and since I wasn't given any inital values I just plug in for 1 into f' and get 6 so that means my answer is 1/6?

4. Originally Posted by alice8675309
Hey thanks so much, mental block there..and since I wasn't given any inital values I just plug in for 1 into f' and get 6 so that means my answer is 1/6?
Well, that's the slope of the tangent, but you need the entire equation of the tangent which is, as I wrote, $t:\, y=\frac{1}{6}(x-3)+1$, or, if you prefer, $t:\, y=\frac{1}{6}x+\frac{1}{2}$.

5. Originally Posted by Failure
Well, that's the slope of the tangent, but you need the entire equation of the tangent which is, as I wrote, $t:\, y=\frac{1}{6}(x-3)+1$, or, if you prefer, $t:\, y=\frac{1}{6}x+\frac{1}{2}$.
Oh yea, I wrote it out like the second one lol I was just making sure it was right, I just forgot to finish on the actual post. But thank you so much!