Derivative of the inverse of f(x) = a polynomial

Question

 

Find the derivative of f^-1(x) at x=6 if f(x) is defined by f(x) = x³ + x.

 

I found the formula g'(x) = 1 / f'(g(x)), where g(x) = f^-1(x). Please explain in detail, because my teacher for AP Calculus AB has only shown us a procedure to find the result, while telling us not to pay attention to use of the formula. It seems like these should be simple problems, but I don't understand these problems very well as a result.

Answer 1

There's no simple way to solve this generally, but we're able to solve it at a given point without too much difficulty - though it does get a little ugly, given the numbers involved here.

 

Based on your formula, you know that we're looking for g'(6) = 1/f'(g(6)), where g(6) = f^-1(6) - i.e., the inverse of f evaluated at six, which is equivalent to the input of f(x) which would yield an output of 6.  That is to say, we're looking for a solution to

 

6 = x³ + x

 

Which we can solve using the cubic formula - thankfully, it's already in depressed cubic form.  We have to find s and t such that

 

3st = 1

s³ - t³ = 6

 

Solving the first for t and plugging into the second gives

 

s³ - 1/27s³ = 6

s⁶ - 6s³ - 1/27 = 0

Which allows s³ to be determined via the quadratic formula as

 

3 ± √(9 + 1/27)

 

So s is

 

(3 ± √(9 + 1/27))^1/3

and t is 1/3 (3 ± √(9 + 1/27))^-1/3

 

Or to turn them into real numbers, s = 1.818 or -.1834

And the corresponding ts are .1834 and -1.818

 

The solutions of such a cubic are s - t, so this gives us 1.634 for x, which we can plug in to verify that it is, in fact, a solution to x³ + x = 6.

So!  Laborious, but done.  Since f(1.634) = 6, it follows that g(6) = 1.634, and we can proceed with the derivative of the inverse.

 

f'(x) = 3x² + 1

 

g'(6) = 1/f'(g(6)) = 1/[3(1.634)² + 1] = .111

 

edit:  Ah, if it's x³ - x, the initial part is much simpler.  x³ - x = 6 has an easy integer solution at x = 2, so g(6) = 2, and g'(6) = 1/f'(2) = 1/[3(2)² - 1] = 1/11