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,

$\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}$

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, ∞)