For all real numbers x find the minimum value of

(x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 14)^2 + (x - 8)^2

Guest Jul 30, 2022

#1**+3 **

differentiating with respect to x (using chain rule) gets:

\(2(x+12)+2(x+7)+2(x+3)+2(x-14)+2(x-8)\\ =2(5x)\\ =10x\)

There is only 1 critical point \(x = 0\), and since this function is concave up (the double derivative is positive), it is a minimum point. Plug in zero to your equation to get your answer.

textot Jul 30, 2022

#1**+3 **

Best Answer

differentiating with respect to x (using chain rule) gets:

\(2(x+12)+2(x+7)+2(x+3)+2(x-14)+2(x-8)\\ =2(5x)\\ =10x\)

There is only 1 critical point \(x = 0\), and since this function is concave up (the double derivative is positive), it is a minimum point. Plug in zero to your equation to get your answer.

textot Jul 30, 2022

#2**0 **

Here's another way, without using calculus:

Expand it to \(5x^2 + 462\)

Now, the minimum value occurs when \(x = 0\), because \(x^2\) is always non-negative.

BuilderBoi Jul 30, 2022