The function$\mathrm{f}\left(\mathrm{x}\right)=4{\mathrm{x}}^3-18{\mathrm{x}}^2+27\mathrm{x}-7$  is always ……… on R.

Given, $\mathrm{f}\left(\mathrm{x}\right)=4{\mathrm{x}}^3-18{\mathrm{x}}^2+27\mathrm{x}-7$
$\mathrm{Differentiating} \mathrm{both} \mathrm{sides} \mathrm{w}.\mathrm{r}.\mathrm{t}. \mathrm{x}, \mathrm{we} \mathrm{get}$
$\begin{array}{r}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=12{\mathrm{x}}^2-36\mathrm{x}+27\\ \Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=12\left({\mathrm{x}}^2-3\mathrm{x}+\frac{9}{4}\right)\\ =12\left[{\left(\mathrm{x}-\frac{3}{2}\right)}^2-\frac{9}{4}+\frac{9}{4}\right]\\ =12{\left(\mathrm{x}-\frac{3}{2}\right)}^2\ge 0\mathrm{\forall }\mathrm{x}\in \mathrm{R}\end{array}$
$\therefore \mathrm{f}\left(\mathrm{x}\right)$ is always increasing on R.