If x∈R, then f(x)=(x−1)2+(x−2)2+(x−3)2+  (x−4)2  assumes its minimum value at

The given equation is f(x)=(x−1)2+(x−2)2+(x−3)2+  (x−4)2 ⇒(x2+1−2x)+(x2+4−4x)+(x2+9−6x)+(x2+16−8x)⇒4x2−2  (1+2+3+4)x+30⇒4x2−20x+30⇒4(x2−5x)+30⇒4  {x2−5x+254}+30−25⇒  4  (x−52)2+5Now,  f(x) is minimum when (x−52)2  is  minimumi.e. when (x−52)2  =0 ;  when x=  52=2.5Alternatively,f(x)  =  (x−a1)2+(x−a2)2+...+(x−an)2Assumes minimum value at  a1+a2+....+annf(x)=(x−1)2+(x−2)2+(x−3)2+  (x−4)2 ⇒  minimum value at  x= 1+2+3+44=104=2.5