If m & M respectively denote the minimum & maximum of  $\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}-1)}^2+3$ for $\mathrm{x}\in [-3,1]$, then the ordered pair (m, M) is equal to

$\begin{array}{c}Given f\left(x\right)={(x-1)}^2+3 for x\in \left[\begin{array}{cc}-3,& 1\end{array}\right]\\ diff w.r.to x on botb sides\\ {\mathrm{f}}^1\left(\mathrm{x}\right)=2(\mathrm{x}-1)=2\mathrm{x}-2\\ for max or min {\mathrm{f}}^1\left(\mathrm{x}\right)=0\\ \Rightarrow 2x-2=0\\ \Rightarrow \mathrm{x}=1\end{array}$

 $$\begin{gathered} \mathrm{Now} \mathrm{f}\left(1\right)=3 \\ \mathrm{f}(-3)=19 \\ \therefore max.value\left(M\right)=19 \\ and min.value\left(m\right)=3 \\ \therefore (m,M)=(3,19) \end{gathered}$$