1. ## Inverse Derivative

For f(x) = 2x^5 + 3x^3 + x

Fi nd (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.

2. 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.

3. Can you please show how you got f^-1(6) = 1

4. 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:
$\displaystyle y=2x^5 + 3x^3 + 1=6$
$\displaystyle 2x^5 + 3x^3 - 5 = 0$
$\displaystyle (x-1)(2x^4+2x^3+5x^2+5x+5)=0$
$\displaystyle x=1$.