Find (f^-1)'(a)

Question

Find (f^-1)'(a)

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

a=0

Answer 1

f(x) = x³ + 2x + 3. So, f'(x) = 3x² + 2

To find [f^-1(0)]', we need to first find the value of x such that f(x) = 0. We see that f(-1) = 0.

Then, [f^-1(0)]' = 1 / [f'(-1)] = 1 / 5

Answer 2

The question calls for the value of the derivative's inverse function at x=0.

First, let's find the derivative of f(x) = x³ + 2x + 3. Using the power rule, we get

f '(x) = 3x² + 2

To find the inverse of a function, we swap the inputs and output. So,

x = 3 ((f^-1)'(x))² +2

(x - 2) / 3 = ((f^-1)'(x))²

(f^-1)'(x) = √( (x - 2) / (3) )

The last step is to evaluate this function at x = a = 0.

(f^-1)'(0) = √( (0 - 2) / (3) )

(f^-1)'(0) = √( -2/3)

We get a non-real answer, the square root of two thirds, so we use the imaginary number i = √(-1).

(f^-1)'(0) = i √(2/3)

Let me know if that makes sense, and if you have any other questions feel free to shoot me a direct message!

-Ethan