It's overall quadratic

Algebra Level 3

The function f ( x ) = ∑ k = 1 5 ( x − k ) 2 f(x)=\displaystyle \sum_{k=1}^5 (x-k)^2 has a minimum value at what value of x x ?

5 2 5 2 \frac 52 3

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Joshua Chin
Nov 3, 2015

For f ( x ) = ∑ k = 1 5 ( x − k ) 2 f\left( x \right) =\sum _{ k=1 }^{ 5 }{ { (x-k) }^{ 2 } }

We obtain

( x − 1 ) 2 + ( x − 2 ) 2 + . . . + ( x − 5 ) 2 = 5 x 2 − ( 2 + 10 2 ) ( 5 ) x + 1 2 + 2 2 + . . . + 5 2 = 5 x 2 − 30 x + 5 × 6 × 11 6 = 5 x 2 − 30 x + 55 = 5 ( x 2 − 6 x + 11 ) = 5 ( x − 3 ) 2 + 2 { (x-1) }^{ 2 }+{ (x-2) }^{ 2 }+...+{ (x-5) }^{ 2 }\\ =5{ x }^{ 2 }-\quad (\frac { 2+10 }{ 2 } )(5)x\quad +\quad { 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ 5 }^{ 2 }\\ =5{ x }^{ 2 }-30x+\frac { 5\times 6\times 11 }{ 6 } \\ =5{ x }^{ 2 }-30x+55\\ =5({ x }^{ 2 }-6x+11)\\ =5{ (x-3) }^{ 2 }+2

Thus

min ⁡ f ( x ) = 2 \min { f(x) } =2 iff

x − 3 = 0 x = 3 x-3=0\\ x=3

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Moderator note:

The minimum of a (standard) parabola occurs at the vertex of − b 2 a - \frac{b}{2a} .

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