Calculus Problem!!! Find the value of df^-1/dx at x = f(a)?

Question

I need assistance with a calculus problem. Specifically, I want to find the value of ( \frac{df^{-1}}{dx} ) at ( x = f(a) ), where ( f(x) = x^3 - 9x^2 - 4 ) and ( x \geq 6 ). I have ( a = 3 ).
I believe the process involves finding the derivative, which I calculated as ( 3x^2 - 18x ). Then, I used the formula ( \frac{1}{f'(f^{-1}(x))} ) to find ( \frac{df^{-1}}{dx} ). 
To clarify my calculations, I found ( f(3) = 3^3 - 9(3^2) - 4 = -58 ). Therefore, I attempted to compute ( \frac{1}{3(-58)^2 - 18(-58)} ), which resulted in ( \frac{1}{11136} ). 
However, this answer seems incorrect, and I'm unsure where I went wrong. Could someone please help me identify the mistake or guide me through the correct process? Thank you!

Anonymous · Accepted Answer

f'(x)=3x^2-18x, then
(df/dx)^-1=1/(3x^2-18x), so
(df/dx)^-1=1/(27-54)=-1/27
But another point to consider is that f^-1 may denote the inverse of f:
f^-1 != 1/f
f^-1 denotes the inverse of f
According to the inverse function theorem:
(d/dx)f^-1(a)=1/f'(b), where a=f(b)
We want (d/dx)f^-1(3), so 3=f(b), so
3=x^3-9x^2-4 -> 0=x^3-9x^2-7
You can use the cubic formula to solve the above eqn. I won't do it for you because it's a pain. Then you plug your solution into (d/dx)f^-1(a)=1/f'(b) 
I don't feel a strong urge to do the analysis, but it appears to me that f^-1 is probably not a function.
Regardless, you'll get 2 complex roots and 1 real root from the cubic formula. The real root is b=9.0848 which gives 1/f'(b)=1/84.074 which may or may not be (d/dx)f^-1(3) depending on whether f^-1 is a function

Madelyn Ebert DDS

Calculus Problem!!! Find the value of df^-1/dx at x = f(a)?

1 Answers

AI-Powered Insights

Want to answer this question?

Categories

Tags

Related Questions