If m and M respectively denote the minimum and maximum values

The given function is $f(x)=(x+1)^{2}+3, x \in[-3,1]$.

$f(x)=(x+1)^{2}+3$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=2(x+1)$

For maxima or minima,

$f^{\prime}(x)=0$

$f^{\prime}(x)=0$

$\Rightarrow 2(x+1)=0$

$\Rightarrow x+1=0$

$\Rightarrow x=-1$

Now, 

$f^{\prime \prime}(x)=2>0$

So, x = −1 is the point of local minimum of f(x).

At x = −1, we have

$f(-1)=(-1+1)^{2}+3=0+3=3$

At x = −3, we have

$f(-3)=(-3+1)^{2}+3=4+3=7$

At x = 1, we have

$f(1)=(1+1)^{2}+3=4+3=7$

Thus, the minimum value of f(x) is 3 and the maximum value of f(x) is 7.

∴ m = 3 and M = 7

Thus, the ordered pair (m, M) is (3, 7).

If m and M respectively denote the minimum and maximum values of f(x) = (x + 1)2 + 3 in the interval [−3, 1], then the ordered pair (m, M) = ___(3, 7)___.