## anonymous 5 years ago Using ONLY calculus to prove that the minimum value of the function f(x)=|x-a1|+|x-a2|+|x-a3|+...+|x-a100| is f(a50) You are given a1<=a2<=a3<=.....<=a100.

1. anonymous

yes..can u do it?

2. watchmath

Nice problem! We will prove in general that the minimum value of $f(x)=\sum_{k=1}^{2n}|x-a_k|$ where $$a_{k}<a_{k+1}$$ for all $$k$$ is $$f(a_n)$$. Just for convenient, let $$a_0=-\infty$$ and $$a_{2n+1}=\infty$$. Notice that for $$x\in(a_k,a_{k+1})$$ we have $$|x-a_i|=x-a_i$$ for all $$0< i\leq k$$ and $$|x-a_i|=-(x-a_i)$$ for $$k< i<2n+1$$. It follows that in $$(a_k,a_{k+1})$$ we have $f(x)=\sum_{i=1}^{k}(x-a_i)+\sum_{i=k+1}^{2n} -(x-a_i)=(2k-2n)x+\sum_{i=k+1}^{2n}a_i-\sum_{i=1}^k a_i$ Note that $$f'(x)=2k-2n$$. So $$f(x)$$ is decreasing on $$(a_k,a_{k+1})$$ for all $$k<n$$ and increasing for $$k> n$$. Hence $$f(x)$$ attain its's minimum value at $$a_n$$.

3. watchmath