1. Let's state the problem: We need to show that the function $g(x) = x^3 - 6x^2 + 18x - 2$ is always increasing. 2. To determine whether $g(x)$ is always increasing, we look at its derivative $g'(x)$ because a function is increasing where its derivative is positive. 3. Compute the derivative: $$ g'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(6x^2) + \frac{d}{dx}(18x) - \frac{d}{dx}(2) = 3x^2 - 12x + 18 $$ 4. We analyze $g'(x)$ to check if it is always positive. This is a quadratic function: $$ g'(x) = 3x^2 - 12x + 18 $$ 5. To determine the sign of $g'(x)$, find its discriminant: $$ \Delta = (-12)^2 - 4 \cdot 3 \cdot 18 = 144 - 216 = -72 $$ 6. Since the discriminant $\Delta$ is negative ($-72 < 0$), the quadratic $g'(x)$ has no real roots and thus does not cross the $x$-axis. 7. Because the leading coefficient $3$ is positive, $g'(x)$ is always positive for all real $x$. 8. Since $g'(x) > 0$ for all $x$, the function $g(x)$ is strictly increasing everywhere. Final answer: $g(x) = x^3 - 6x^2 + 18x - 2$ is strictly increasing for all real $x$ because its derivative $g'(x) = 3x^2 - 12x + 18$ is always positive.