Show that the function $f$ given by $f(x) = x^{3} - 3x^{2} + 4x$,$x \in R$ is strictly increasing on $R$.

(N/A) To determine if the function $f(x) = x^{3} - 3x^{2} + 4x$ is increasing on $R$,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{3} - 3x^{2} + 4x) = 3x^{2} - 6x + 4$
Now,we rewrite the expression for $f'(x)$ by completing the square:
$f'(x) = 3(x^{2} - 2x) + 4$
$f'(x) = 3(x^{2} - 2x + 1 - 1) + 4$
$f'(x) = 3((x - 1)^{2} - 1) + 4$
$f'(x) = 3(x - 1)^{2} - 3 + 4$
$f'(x) = 3(x - 1)^{2} + 1$
Since $(x - 1)^{2} \geq 0$ for all $x \in R$,it follows that $3(x - 1)^{2} \geq 0$.
Adding $1$ to both sides,we get $3(x - 1)^{2} + 1 \geq 1$.
Thus,$f'(x) > 0$ for all $x \in R$.
Since the derivative $f'(x)$ is strictly greater than $0$ for all $x$ in the domain $R$,the function $f(x)$ is strictly increasing on $R$.