What is the nature of function f(x) = x3 – 3x2 + 4x on R?

The correct option is Increasing.

To determine the nature of the function $f\left(x\right)=x^3-3x^2+4x$ on the set of real numbers ($R$), we can analyze its first derivative with respect to $x$. The first derivative, $f^{\prime }\left(x\right)$, represents the rate of change of the function and helps us determine whether the function is increasing or decreasing.

First, let's find the derivative of $f\left(x\right)$:
$$f^{\prime }\left(x\right)=\frac{d}{dx}(x^3-3x^2+4x)$$
Applying the power rule of differentiation, we get:
$$f^{\prime }\left(x\right)=3x^2-6x+4$$

Next, we analyze the sign of the quadratic expression $3x^2-6x+4$. We can write this expression by completing the square:
$$f^{\prime }\left(x\right)=3(x^2-2x)+4$$
$$f^{\prime }\left(x\right)=3(x^2-2x+1-1)+4$$
$$f^{\prime }\left(x\right)=3{(x-1)}^2-3+4$$
$$f^{\prime }\left(x\right)=3{(x-1)}^2+1$$

Since the square of any real number is always non-negative (i.e., ${(x-1)}^2\ge 0$ for all $x\in R$), it follows that:
$$3{(x-1)}^2\ge 0$$
Adding 1 to both sides, we obtain:
$$3{(x-1)}^2+1\ge 1>0$$

Since $f^{\prime }\left(x\right)>0$ for all real values of $x$, the slope of the function is strictly positive everywhere on $R$. Therefore, the function $f\left(x\right)$ is strictly increasing on $R$.