#Mark the correct alternative in each of the following
The function assumes minimum value at x =
f(x)=(x-1)2+(x-2)2+(x-3)2+(x-4)2+(x-5)2
f’(x)=2[5x-15]
f’(x)=0;x=3
Hence by second derivative test
f’’(x)>0 so it’s a point of minimum.
f”(x)=1>0 so At x=3 minimum value.Option(C)