For f(x) = 2x^5 + 3x^3 + x
Find (f^-1)' (6)
How do you do this?
The formula is d/dx (f^-1(x)) = 1/ (f'(f^-1(x))
How do I set it up?
Thanks.
As soon as you're given f(x), you can find its derivative $\displaystyle f'(x)=10x^{4}+9x^{2}+1$. You also need to find the inverse of f evaluated at 6: $\displaystyle f^{-1}(6)=1$ by inspection.
Now you just plug that information into your equation $\displaystyle \left.\begin{matrix}\frac{d}{dx}f^{-1}(x) \end{matrix}\right|_{x=6}=\frac{1}{f'(f^{-1}(6))}$ to get the answer.
To find $\displaystyle f^{-1}(6)$, you need to find a value for x that makes y equal to 6. That is, you need to find the solution for $\displaystyle 2x^5 + 3x^3 + 1 = 6$.
Notice that the coefficients of f(x) add up to 6. Knowing this, all you have to do is plug in x=1 to make the x's "disappear":
$\displaystyle 2(1)^5 + 3(1)^3 + 1 = 2 + 3 + 1 = 6$.
A general way to solve this type of problem is to use synthetic division to factor the polynomial:
$\displaystyle
y=2x^5 + 3x^3 + 1=6
$
$\displaystyle
2x^5 + 3x^3 - 5 = 0
$
$\displaystyle
(x-1)(2x^4+2x^3+5x^2+5x+5)=0
$
$\displaystyle
x=1
$.