The function \(f(x)=x^3-3 x^2+12 x-18\) is :

Given: \(f(x)={x}^{3}-3{x}^{2}+12x-18\)

Then, \({f}^{'}(x)=3{x}^{2}-6x+12=3\left({x}^{2}-2x+4\right)\)

since, \({x}^{2}-2x+4=(x-1{)}^{2}+3>0\forall x\in \mathrm{ℝ}\)

\(\Rightarrow {f}^{'}(x)>0\forall x\in \mathrm{ℝ}\)

\(\Rightarrow f\) is strictly increasing on \(\mathbb{R}\)