The smallest value of ${\mathrm{x}}^2-3\mathrm{x}+3$  in the interval $\left[\begin{array}{cc}-3,& \frac{3}{2}\end{array}\right]$  is

$$\begin{gathered} \mathrm{Given} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-3\mathrm{x}+3 \\ \mathrm{diff} \mathrm{w},\mathrm{r}.\mathrm{to} \mathrm{x} \mathrm{on} \mathrm{both} \mathrm{sides} \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=2\mathrm{x}-3 \\ f\text{'}\text{'}\left(x\right)=2 \\ for max or min {\mathrm{f}}^1\left(\mathrm{x}\right)=0 \\ 2x-3=0\Rightarrow \mathrm{x}=\frac{3}{2} \\ clearly f\text{'}\text{'}\left(x\right)>0 \\ Now f(-3)=21 \\ \mathrm{f}\left(\frac{3}{2}\right)=\frac{3}{4} \\ Now smallest value of f\left(x\right)=\mathrm{Minimum} \left\{\mathrm{f}(-3),\mathrm{f}\left(\frac{3}{2}\right)\right\}=\frac{3}{4} \end{gathered}$$