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(B) Given the function $f(x) = (x-1)^2 + 3$ defined on the interval $x \in [-3, 1]$.First,we find the critical points by differentiating $f(x)$ with r

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$(3, 19)$

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(B) Given the function $f(x) = (x-1)^2 + 3$ defined on the interval $x \in [-3, 1]$.First,we find the critical points by differentiating $f(x)$ with respect to $x$:$f'(x) = 2(x-1)$.Setting $f'(x) = 0$ gives $2(x-1) = 0$,which implies $x = 1$.Since $x = 1$ is an endpoint of the given interval $[-3, 1]$,we evaluate the function at the critical point and the boundaries of the interval:At $x = 1$: $f(1) = (1-1)^2 + 3 = 3$.At $x = -3$: $f(-3) = (-3-1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$.Comparing these values,the minimum value $m = 3$ and the maximum value $M = 19$.Therefore,the ordered pair $(m, M)$ is $(3, 19)$.

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

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