Given that, f(x)=(x−1)2+3,‌‌x∈[−3,1] On differentiating w.r.t. x, we get f′(x)=2(x−1) For maxima and minima, put f′(x)=0 ⇒‌2(x−1)‌=0 ⇒‌x‌=1 Now, f′′(x)=2, minima ∀x∈R At ‌‌x=1, f(1)=(1−1)2+3=3 At ‌‌x=−3, f(−3)=(−3−1)2+3=19 Here, m=3 and M=19 Hence, required ordered pair is (3,19).
