Question

If f(x)=x^3-x+2, then (f^-1)'(2)

Answer 1

Note that f(x) as given is not invertible. By definition of inverse function,

`f\left(f^(-1)(x)\right) = x`

`\implies f^(-1)(x)^3 - f^(-1)(x) + 2 = x`

which is a cubic polynomial in
`f^(-1)(x)` with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

`\left(f^(-1)\right)'(b) = \frac1{f'(a)}`

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0 or x = 1 or x = -1

Then
`\left(f^(-1)\right)'(2)` can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)