The problem asks us to find the value(s) of $a$ such that the function $f(x) = x^3 + ax^2 + 3x - 4$ is always increasing.

1. Problem Description

The problem asks us to find the value(s) of $a$ such that the function $f(x) = x^3 + ax^2 + 3x - 4$ is always increasing.

2. Solution Steps

For the function $f(x)$ to be always increasing, its derivative $f'(x)$ must be greater than or equal to 0 for all $x$.

First, we find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(x^3 + ax^2 + 3x - 4) = 3x^2 + 2ax + 3$.

Now we need to ensure that $f'(x) \ge 0$ for all $x$. This means that the quadratic $3x^2 + 2ax + 3$ should either have no real roots or one real root. In other words, its discriminant must be less than or equal to zero.

The discriminant of a quadratic $Ax^2 + Bx + C$ is given by $B^2 - 4AC$.

In our case, $A = 3$, $B = 2a$, and $C = 3$. Thus, the discriminant is

$(2a)^2 - 4(3)(3) = 4a^2 - 36$.

We require the discriminant to be less than or equal to 0:

$4a^2 - 36 \le 0$.

$4a^2 \le 36$.

$a^2 \le 9$.

$-3 \le a \le 3$.

3. Final Answer

The values of $a$ for which the function $f(x)$ is always increasing are $-3 \le a \le 3$.