Suppose f(x) = 3x + 4. What is f^-1(x) and what is (f^-1)'(x).
f^-1(x) = (x-4)/3
so, the derivative is 1/3
To find the inverse of a function, f(x), we need to interchange the roles of x and y in the function and solve for y. In this case, we have:
f(x) = 3x + 4
Step 1: Interchange x and y:
x = 3y + 4
Step 2: Solve for y:
x - 4 = 3y
y = (x - 4)/3
So, the inverse function of f(x) = 3x + 4 is f^(-1)(x) = (x - 4)/3.
Now, to find the derivative of the inverse function, (f^(-1))'(x), we can use the inverse function theorem, which states that if f(x) and f^(-1)(x) are inverses, then the derivative of f^(-1)(x) is equal to 1 divided by the derivative of f(x) evaluated at f^(-1)(x).
Step 1: Find the derivative of f(x):
f'(x) = 3
Step 2: Evaluate the derivative of f(x) at f^(-1)(x):
[f^(1)]'(x) = 1/f'(f^(-1)(x))
= 1/3 (since f'(x) = 3 for all x)
Therefore, the derivative of f^(-1)(x) is (f^(-1))'(x) = 1/3.