Find (f −1)'(a). f(x) = 5x3 + 4x2 + 8x + 2, a = 2

Question

I'm confused on how to solve this one with y^2 and y in the same equation.

Answer 1

Looks like it is asking for the inverse function of f(x) at x=2.

You could find the inverse of the function first, however, there is a formula for finding the derivative of the inverse function:

d/dx[f^-1(a)] = 1/f'(b)

The value of b is found by setting the original f(x) to a and solving for x

2 = 5x³+4x²+8x+2

x = 0 by observation. Other roots are complex. So b = 0.

f '(x) = 15x²+8x+8

f'(0) = 8

1/f'(0) = 1/8

d/dx[f^-1(2)] = 1/8

Answer 2

By inspection, f(0) = 2

So, the point (0,2) is on the graph of y = f(x) and the point (2,0) is on the graph of the inverse function.

To find the derivative of f^-1(x) at the point (2,0), take the reciprocal of the derivative of f(x) at the point (0,2).

In other words, the slope of the tangent to the graph of y = f^-1(x) at (2,0) is the reciprocal of the slope of the tangent to the graph of y = f(x) at the point (0,2).

f'(x) = 15x² + 8x + 8

Slope of tangent to the graph of y = f(x) at (0,2) is f'(0) = 8.

Therefore, the slope of the tangent (the derivative of f^-1(x)) when x = 2 is 1/8.