函数 $\text{f}$ 对所有实数 $x$ 有定义，且满足 $\text{f}''(x) = 6x + 2$。
曲线 $y = \text{f}(x)$ 在点 $(-1, 10)$ 处的斜率为 $11$。

1. Integrate the second derivative to find the general form of the first derivative.

$\text{f}'(x) = \int (6x + 2) \, dx$
$\text{f}'(x) = \frac{6x^2}{2} + 2x + c$
$\text{f}'(x) = 3x^2 + 2x + c$

2. Use the given gradient at $x = -1$ to find the constant of integration $c$.

Given that the gradient to the curve at $(-1, 10)$ is $11$, we have $\text{f}'(-1) = 11$.
$3(-1)^2 + 2(-1) + c = 11$
$3 - 2 + c = 11$
$1 + c = 11$
$c = 10$

3. State the final expression for $\text{f}'(x)$.

$\text{f}'(x) = 3x^2 + 2x + 10$

1. Find the expression for f'(x) from part (a)

$\text{f}''(x) = 6x + 2$
$\text{f}'(x) = \int (6x + 2) \, dx = 3x^2 + 2x + c$
Given $\text{f}'(-1) = 11$:
$3(-1)^2 + 2(-1) + c = 11$
$3 - 2 + c = 11 \implies c = 10$
$\text{f}'(x) = 3x^2 + 2x + 10$

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

$\text{f}(x) = \int (3x^2 + 2x + 10) \, dx$
$\text{f}(x) = x^3 + x^2 + 10x + d$

3. Use the given point (-1, 10) to find the constant of integration d

Since the curve passes through $(-1, 10)$:
$10 = (-1)^3 + (-1)^2 + 10(-1) + d$
$10 = -1 + 1 - 10 + d$
$10 = -10 + d \implies d = 20$
$\text{f}(x) = x^3 + x^2 + 10x + 20$

1. State the expression for $\text{f}'(x)$ found in part (a) and set it to zero to find stationary points

From part (a), $\text{f}'(x) = 3x^2 + 2x + 10$.
For stationary points, $\text{f}'(x) = 0$:
$3x^2 + 2x + 10 = 0$

2. Calculate the discriminant of the quadratic equation

Discriminant, $\Delta = b^2 - 4ac$
$\Delta = (2)^2 - 4(3)(10)$
$\Delta = 4 - 120 = -116$

3. Conclude whether stationary points exist based on the discriminant

Since $\Delta < 0$, the equation $3x^2 + 2x + 10 = 0$ has no real roots.
Therefore, $\text{f}'(x) \neq 0$ for all real $x$, meaning the curve has no stationary points.