+0  
 
0
109
2
avatar

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

 Jul 30, 2022

Best Answer 

 #1
avatar+502 
+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.

 Jul 30, 2022
 #1
avatar+502 
+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
avatar+2448 
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.

 Jul 30, 2022

23 Online Users