Let f(x) be a cubic polynomial where f(2) = 6 and f(-1) = 15. Also, f(x) takes on stationary values at x = -2 and x = 1. Find f(x).

1. Express the derivative $f'(x)$ using the given stationary points.

Since $f(x)$ is a cubic polynomial, its derivative $f'(x)$ is a quadratic polynomial. Stationary values occur at $x = -2$ and $x = 1$, so these are the roots of $f'(x) = 0$.
$f'(x) = k(x+2)(x-1)$
$f'(x) = k(x^2 + x - 2)$

2. Integrate $f'(x)$ to find the general form of $f(x)$.

$f(x) = \int k(x^2 + x - 2) \, dx$
$f(x) = k\left( \frac{x^3}{3} + \frac{x^2}{2} - 2x \right) + c$

3. Substitute the given points to form a system of simultaneous equations.

Substitute $f(2) = 6$:
$k\left( \frac{8}{3} + \frac{4}{2} - 4 \right) + c = 6$
$\frac{2}{3}k + c = 6 \quad \text{--- (1)}$

Substitute $f(-1) = 15$:
$k\left( -\frac{1}{3} + \frac{1}{2} - 2(-1) \right) + c = 15$
$\frac{13}{6}k + c = 15 \quad \text{--- (2)}$

4. Solve the equations for $k$ and $c$, then write the final expression for $f(x)$.

Subtract (1) from (2):
$\left( \frac{13}{6} - \frac{4}{6} \right)k = 15 - 6$
$\frac{9}{6}k = 9 \implies k = 6$

Substitute $k = 6$ into (1):
$\frac{2}{3}(6) + c = 6$
$4 + c = 6 \implies c = 2$

Thus,
$f(x) = 6\left( \frac{x^3}{3} + \frac{x^2}{2} - 2x \right) + 2$
$f(x) = 2x^3 + 3x^2 - 12x + 2$