函数 $\mathrm{f}(x)$ 是一个四次多项式，在 $A(1, 5)$ 和 $B(4, 0)$ 处有驻点。
当 $x > 4$ 时，$\mathrm{f}(x)$ 是递增函数。
当 $x < 4$ 时，$\mathrm{f}(x)$ 不是递增函数。

1. Determine the nature of stationary point B

Since $\mathrm{f}(x)$ is an increasing function when $x > 4$, $\mathrm{f}'(x) > 0$ for $x > 4$.
Since $\mathrm{f}(x)$ is not an increasing function when $x < 4$, and $B(4,0)$ is a stationary point, $\mathrm{f}'(x) < 0$ for $x$ just before 4.
Thus, the gradient changes from negative to positive at $x = 4$.
Therefore, $B$ is a minimum point.

2. Determine the nature of stationary point A

We are given that $\mathrm{f}(x)$ is not an increasing function when $x < 4$. This means $\mathrm{f}'(x) \leqslant 0$ for $x < 4$.
Since $A(1,5)$ is a stationary point, $\mathrm{f}'(1) = 0$.
For $x < 1$, $\mathrm{f}'(x) < 0$.
For $1 < x < 4$, $\mathrm{f}'(x) < 0$.
The gradient does not change sign at $x = 1$ (it is negative on both sides).
Therefore, $A$ is a stationary point of inflexion.

1. Expand the expression for $\mathrm{f}'(x)$ to prepare for integration.

$\mathrm{f}'(x) = a(x^2 - 2x + 1)(x - 4)$
$\mathrm{f}'(x) = a(x^3 - 4x^2 - 2x^2 + 8x + x - 4)$
$\mathrm{f}'(x) = a(x^3 - 6x^2 + 9x - 4)$

2. Integrate $\mathrm{f}'(x)$ to find the general form of $\mathrm{f}(x)$.

$\mathrm{f}(x) = \int a(x^3 - 6x^2 + 9x - 4) \, dx$
$\mathrm{f}(x) = a\left(\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} - 4x\right) + c$

3. Use the given stationary points $A(1, 5)$ and $B(4, 0)$ to form simultaneous equations for $a$ and $c$.

Substitute $x = 1, \mathrm{f}(x) = 5$:
$5 = a\left(\frac{1}{4} - 2 + \frac{9}{2} - 4\right) + c$
$5 = a\left(-\frac{5}{4}\right) + c \implies c = 5 + \frac{5}{4}a$

Substitute $x = 4, \mathrm{f}(x) = 0$:
$0 = a\left(\frac{4^4}{4} - 2(4^3) + \frac{9(4^2)}{2} - 4(4)\right) + c$
$0 = a(64 - 128 + 72 - 16) + c$
$0 = a(-8) + c \implies c = 8a$

4. Solve the simultaneous equations to find $a$ and $c$.

$8a = 5 + \frac{5}{4}a$
$\frac{27}{4}a = 5 \implies a = \frac{20}{27}$
$c = 8\left(\frac{20}{27}\right) = \frac{160}{27}$

5. Write the final expression for $\mathrm{f}(x)$.

$\mathrm{f}(x) = \frac{20}{27}\left(\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} - 4x\right) + \frac{160}{27}$
$\mathrm{f}(x) = \frac{5}{27}x^4 - \frac{40}{27}x^3 + \frac{30}{27}x^2 - \frac{80}{27}x + \frac{160}{27}$