Question

Prove that the minimum value of (x-a)^2 + (x-b)^2 occurs when x=a+b/2. what is the minimum value?

Answer 1

The function f(x)= (x-a)^2 has a minimum value at that point where f(x)=0 and if the solution of f(x)=0 is x=c, f'(c) is positive. f(x)= 2*(x-a)+2*(x-b) f'(x)=2+2=4 which is always positive f(x)=0 => 2*(x-a)+2*(x-b) => 2x-2a+2x-2b