Prompt:
Given $a, b, c, d \in R$
Find a function $f(x)$ such that:
$f'(a) = 0$
$f'(c) = 0$
$f(a) = b$
$f(c) = d$
Solution
Cubic form is $k \cdot g(x) + h$
For $f'(a)$ and $f'(c)$ to be equal to $0$, $a$ and $c$ must be the roots of a polynomial:
$f'(x) = (x - a)(x - c)$
Integrate:
$g(x) = \int (x - a)(x - c) dx$
$g(x) = \int (x^2 - ax - cx + ac) dx$
$g(x) = \int x^2 dx - \int ax dx - \int cx dx + \int ac dx$
Factor out constants:
$g(x) = \int x^2 dx - a\int x dx - c\int x dx + ac\int 1 dx$
$g(x) = \frac{x^3}{3} - \frac{ax^2}{2} - \frac{cx^2}{2} + acx$
$g(x) = \frac{x^3}{3} - \frac{ax^2 - cx^2}{2} + acx$
$g(x) = \frac{x^3}{3} - \frac{x^2(a-c)}{2} + acx$
Factor out $x$:
$g(x) = x(\frac{x^2}{3} - \frac{x(a-c)}{2} + ac)$
$g(x) = \frac{6x(\frac{x^2}{3} - \frac{x(a-c)}{2} + ac)}{6}$
$g(x) = \frac{x(2x^2 - 3x(a-c) + 6ac)}{6}$
Return to original equation: $kx(\frac{x^2}{3} - \frac{x(a-c)}{2} + ac) + h$
Note that:
$f(a) = b$
$ka(\frac{a^2}{3} - \frac{a(a-c)}{2} + ac) + h = b$
$k(\frac{a^3}{3} - \frac{a^2(a-c)}{2} + a^2c) + h = b$
$f(c) = d$
$kc(\frac{c^2}{3} - \frac{c(a-c)}{2} + ac) + h = d$
$k(\frac{c^3}{3} - \frac{c^2(a-c)}{2} + ac^2) + h = d$
Solve system of equations by subtraction:
$b - d = f(a) - f(c)$
$b - d = (k(\frac{a^3}{3} - \frac{a^2(a-c)}{2} + a^2c) + h) - (k(\frac{c^3}{3} - \frac{c^2(a-c)}{2} + ac^2) + h)$
Ignoring $h$ as it does not change the value of $k$ in any way:
$b - d = (k(\frac{a^3}{3} - \frac{a^2(a-c)}{2} + a^2c)) - (k(\frac{c^3}{3} - \frac{c^2(a-c)}{2} + ac^2))$
Factor out $k$.
$b - d = k((\frac{a^3}{3} - \frac{a^2(a-c)}{2} + a^2c) - (\frac{c^3}{3} - \frac{c^2(a-c)}{2} + ac^2))$
Factor out $-1$ from second term.
$b - d = k(\frac{a^3}{3} - \frac{a^2(a-c)}{2} + a^2c - \frac{c^3}{3} - \frac{c^2(a-c)}{2} - ac^2)$
Group fractions.
$b - d = k(\frac{a^3 - c^3}{3} - \frac{a^2(a-c) - c^2(a - c)}{2} + ac(a - c))$
$b - d = k(\frac{a^3 - c^3}{3} - \frac{(a^2 - c^2)(a - c)}{2} + ac(a - c))$
$b - d = k(\frac{(a - c)(a^2 + ac + c^2)}{3} - \frac{(a^2 - c^2)(a - c)}{2} + ac(a - c))$
$b - d = k(a - c)(\frac{a^2 + ac + c^2}{3} - \frac{a^2 - c^2}{2} + ac)$
$b - d = k(a - c)(\frac{a^2 + ac + c^2}{3} - \frac{(a + c)(a + c)}{2} + ac)$
[DNF] but,
$k = -\frac{6(b - d)}{(a - c)^3}$
$h = \frac{b + d}{2} - \frac{(b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3}$
Finally,
$f(x) = -\frac{6(b - d)}{(a - c)^3}(\frac{x(2x^2 - 3x(a-c) + 6ac)}{6}) + (\frac{b + d}{2} - \frac{(b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3})$
$f(x) = -\frac{(b - d)}{(a - c)^3}(x(2x^2 - 3x(a-c) + 6ac) + (\frac{b + d}{2} - \frac{(b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3})$
$f(x) = -\frac{(b - d)(x(2x^2 - 3x(a-c) + 6ac)}{(a - c)^3} + \frac{b + d}{2} - \frac{(b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3}$
$f(x) = -\frac{(b - d)(x(2x^2 - 3x(a-c) + 6ac)}{(a - c)^3} + \frac{(b + d)(a - c)^3}{2(a - c)^3} - \frac{(b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3}$
$f(x) = -\frac{(b - d)(x(2x^2 - 3x(a-c) + 6ac)}{(a - c)^3} + \frac{(b + d)(a - c)^3 - (b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3}$
$f(x) = -\frac{2(b - d)(x(2x^2 - 3x(a-c) + 6ac)}{2(a - c)^3} + \frac{(b + d)(a - c)^3 - (b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3}$
$f(x) = -\frac{2(b - d)(x(2x^2 - 3x(a-c) + 6ac) + (b + d)(a - c)^3 - (b - d)(a + c)(a^2 + c^2 - 4ac)}{2(a - c)^3}$
