Answer: To find the value of (f^-1)'(x) when x=3, we need to use the formula for the derivative of the inverse function:
(f^-1)'(x) = 1 / f'(f^-1(x))
Here, f(x) = (1/4)x^3 + x - 1. To find f'(x), we need to take the derivative of f(x) with respect to x:
f'(x) = (3/4)x^2 + 1
Now, we need to find f^-1(x) by solving for x in terms of y:
y = (1/4)x^3 + x - 1
y + 1 = (1/4)x^3 + x
4(y + 1) = x^3 + 4x
x^3 + 4x - 4(y + 1) = 0
We can solve this equation for x using a numerical method such as Newton's method or by using a graphing calculator. After solving, we find that:
f^-1(x) = g(x) = cube root of (x - 3) + 1
Now, we can find (f^-1)'(x) when x = 3:
(f^-1)'(3) = 1 / f'(f^-1(3))
= 1 / f'(g(3))
= 1 / [(3/4)g(3)^2 + 1]
= 1 / [(3/4)(-1)^2 + 1]
= 1 / (3/4 + 1)
= 4/7
Therefore, the value of (f^-1)'(x) when x = 3 is 4/7.
Explanation: