Let f(x) = x^3+5x+12

Question

a) Find (f^-1)'(18)

c) Find the equation of the tangent line to the curve y = f^-1(x) when x=18

Answer 1

First calculate f', which will be useful for a couple purposes.

f'(x) = 3x² + 5

Please note that for all x, 3x² + 5 > 0, and therefore f(x) is monotone increasing,

therefore f^-1 is indeed a function (not a mere relation).

Now calculate x = f^-1(18).

It is the value x satisfying

f(x) = 18

x³ + 5x + 12 = 18

x³ + 5x - 6 = 0

Just by observation x=1 is a solution.

To calculate (f^-1)'(18) recall that f^-1 is a reflection of f around the 45° line, therefore

(f^-1)'(18) = 1/f'(1) = 1/8

That is the answer to question a)

b)

To avoid any confusion let me point out that up to now we were using x as the

independent variable of f; from now on (following the statement of the problem)

x is the independent variable of f^-1.

The tangent is the line with slope 1/8 passing through the point (18, 1).

That is

y = 1/8(x-18) + 1