Inverse Function:
there is a formula that says: $\displaystyle \left(f^{-1} \right)'(a) = \frac {1}{f' \left( f^{-1}(a) \right)}$ ...i can show you how to derive this if you are interested.
can you continue? the only problem here really, is finding $\displaystyle f^{-1}(a)$, do you think you can do that?
So we know $\displaystyle \left(f^{-1} \right)'(a) = \frac {1}{f' \left( f^{-1}(a) \right)}$
and we wish to find $\displaystyle \left(f^{-1} \right)'(1) $ for the function $\displaystyle f(x) = x^3 + x + 1$
You know that $\displaystyle f'(x) = 3x^2 + 1$
Now we need to find $\displaystyle f^{-1}(1)$
Recall that the inverse function is a function that takes an output value back to its preimage (or input value). So saying we want to find $\displaystyle f^{-1}(1)$ is saying that we want to find an $\displaystyle x$ such that $\displaystyle f(x) = 1$
Now I don't really know any shortcuts here, it's just something you have to figure out by working backwards in your head. usually they give you functions where it is easy to tell what $\displaystyle x$ gives you the desired value. Here it is obviously 0, since $\displaystyle f(0) = 0^3 + 0 + 1 = 1$. Therefore $\displaystyle f^{-1}(1) = 0$ since $\displaystyle f(0) = 1$
So now, finally, we plug it into our formula:
$\displaystyle \left(f^{-1} \right)'(1) = \frac {1}{f' \left( f^{-1}(1) \right)}$
..............$\displaystyle = \frac {1}{f' \left(0 \right)}$
..............$\displaystyle = \frac {1}{3(0)^2 + 1}$
..............$\displaystyle = 1$
Jhevon you are the freaking MAN!! A true Jedi Master!!
OK so let me get this straight...
1) Take the derivative of the function
2) Set the function = to that "a" value and solve for X
3) 1/derivative and place the "X" value from step 2 where x is in the (f^-1)'(a) formula?
Is this correct?
Thanks Buddy!
-qbkr21