f(x)=Σ_(k=1) ^n ∣x+k∣ (1) find the values of x such that f(x) is minumum. (2) fund the roots of f(x)−m=0 as example you can set n=100, m=2500.


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Question Number 69276 by mr W last updated on 22/Sep/19

$f (x) = \underset{k = 1}{\sum^{n}} ∣ x + k ∣$ $(1) f i n d t h e v a l u e s o f x s u c h t h a t f (x)$ $i s m i n u m u m .$ $(2) f u n d t h e r o o t s o f f (x) - m = 0$ $a s e x a m p l e y o u c a n s e t n = 100, m = 2500 .$
Commented by Prithwish sen last updated on 22/Sep/19

$I tink$ $f (x) will be minimum if we shift the origin$ $to the - A$ $where A = mean of 1, 2, 3, 4 . . . . . . . . . n$ $∴ x = - \frac{n + 1}{2}$ $f (x) = ∣ - \frac{n + 1}{2} + 1 ∣ . . . . . . . . . . . . . ∣ - \frac{n + 1}{2} + n ∣$ $=$ $Now if we take n = 100$ $then f (x) = 99 + 97 + 95 + . . . . . . . 5 + 3 + 1 = 50^{2} = 2500$ $2) Now the root of f (x) for m = 2500$ $f (- 50.5) - 2500 = 0$ $i . e the root of f (x) - m is x = - 50.5$
Commented by Prithwish sen last updated on 23/Sep/19

$Thank you sir . I was egarly waiting for your$ $valuable suggestion .$
Commented by mr W last updated on 22/Sep/19

$t h a n k s a l o t s i r!$ $y o u a r e g e n e r a l l y r i g h t . b u t w e c a n b e$ $m o r e e x a c t :$ $s i n c e t h e c u r v e o f f (x) i s a k i n d o f$ $f o l d e d l i n e, w e c a n s a y :$ $i f n i s o d d, t h e n f {(x)}_{m i n} i s a t x = - \frac{n + 1}{2},$ $a n d i f n i s e v e n, t h e n f {(x)}_{m i n} i s w h e n$ $x \in [- (\frac{n}{2} + 1), - \frac{n}{2}] .$
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