If f and h are inverses of one another, then we have
h(f(x)) = x
Differentiating both sides using the chain rule gives
h'(f(x)) • f'(x) = 1
Let f(x) = -1. Then it would follow that
h'(-1) = 1/f'(x)
for the value of x such that f(x) = -1. This result is known as the inverse function theorem.
Solve for x such that f(x) = -1 :
2x³ - 3x = -1
2x³ - 3x + 1 = 0
By inspection, x = 1 is a solution, since 2 - 3 + 1 = 0. So x - 1 is a factor of the left side; polynomial division yields a factorization of
(x - 1) (2x² + 2x - 1) = 0
and using the quadratic formula, we find three solutions,
x = 1, x = (-1 - √3)/2, x = (-1 + √3)/2
Since there's more than one solution here, this means f(x) is not one-to-one and thus has no inverse in the strict sense of the term.
We can however restrict the domain of f(x) to a subset over which it can be inverted. Using the fact that polynomials are monotonic over intervals between their extrema, we find the critical points of f(x) :
f'(x) = 6x² - 3 = 0 ⇒ x² = 1/2 ⇒ x = ± 1/√2
Check the sign of the second derivative at each critical point:
f''(x) = 12x
• f''(-1/√2) = -12/√2 < 0 ⇒ x = -1/√2 is a local maximum
• f''(1/√2) = 12/√2 > 0 ⇒ x = 1/√2 is a local minimum
Now we have a few choices of how to define the inverse of f(x).
• If we restrict the domain of f(x) to the interval (-∞, -1/√2), then f(x) = -1 when x = (-1 - √3)/2. Then by the inverse function theorem,
f((-1 - √3)/2) = -1 ⇒ h'(-1) = 1/f'((-1 - √3)/2) = (-1 + √3)/6
• If we instead restrict the domain to (-1/√2, 1/√2), then f(x) = -1 when x = (-1 + √3)/2. Then
f((-1 + √3)/2) = -1 ⇒ h'(-1) = 1/f'((-1 + √3)/2) = (-1 - √3)/6
• Otherwise, if we restrict to (1/√2, ∞), then f(x) = -1 when x = 1. Then
f(1) = -1 ⇒ h'(-1) = 1/f'(1) = 1/3