SUMO-5

Algebra Level 2

Compute the minimum possible value of: ( x − 1 ) 2 + ( x − 2 ) 2 + ( x − 3 ) 2 + ( x − 4 ) 2 + ( x − 5 ) 2 (x-1)^2+(x-2)^2+(x-3)^{2}+(x-4)^2+(x-5)^2 for real values of x x .

SUMO: Stanford University Math Organization


The answer is 10.

Hana Wehbi by Hana Wehbi , 51, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Samrit Pramanik
Jun 13, 2018

Let f ′ ( x ) = ( x − 1 ) 2 + ( x − 2 ) 2 + ( x − 3 ) 3 + ( x − 4 ) 2 + ( x − 5 ) 2 f'(x)=(x-1)^2+ (x-2)^2+(x-3)^3+(x-4)^2+(x-5)^2

∴ f ′ ( x ) = 2 ( x − 1 ) + 2 ( x − 2 ) + 3 ( x − 3 ) + 2 ( x − 4 ) + 2 ( x − 5 ) = 11 x − 33 \therefore f'(x)=2(x-1)+2(x-2)+3(x-3)+2(x-4)+2(x-5) = 11x-33

and f ′ ′ ( x ) = 11 > 0 f''(x)=11>0

f ′ ( x ) = 0 ⇒ x = 3 f'(x)=0 \Rightarrow x=3

So, the minimum value of f ( x ) f(x) is f ( 3 ) f(3) i.e. 10 10

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that it is ( x − 3 ) 3 (x-3)^{\color{#D61F06}3} and not ( x − 3 ) 2 (x-3)^{\color{#3D99F6}2} but the answer is the same.

Chew-Seong Cheong - 2 years, 12 months ago

Log in to reply

oooopsss....i am really sorry..i could not notice..

Samrit Pramanik - 2 years, 12 months ago

Ok..now I've edited the solution..Thank you

Samrit Pramanik - 2 years, 12 months ago

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...