1. f(x) = 2x^3 -3x. If h(x) is the inverse function of f, then h'(-1) = I believe the correct answer is 1/3, but I don't know how to get it.

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