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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The function f(x) = (x - a)^2 + (x - b)^2 has a minimum value at the 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) = 0

=> 2x - 2a + 2x - 2b = 0

=> 4x = 2a + 2b

=> x = `(a + b)/2`

The minimum value of (x - a)^2 + (x - b)^2 occurs at x = `(a + b)/2` .

`f((a + b)/2) = (a/2 + b/2 - a)^2 + (a/2 + b/2 - b)^2`

=> `(b/2 - a/2)^2 + (a/2 - b/2)^2`

=> `(1/4)(b^2 + a^2 - 2ab + a^2 + b^2 - 2ab)`

=> `(a - b)^2/2`

The minimum value of (x - a)^2 + (x - b)^2 occurs at `x = (a + b)/2` and is equal to `(a - b)^2/2`

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