Difference between revisions of "MIE 2016/Day 1/Problem 8"
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==Solution == | ==Solution == | ||
| + | The minimum value of this function is when <math>x</math> is in the middle between 1 and 2017. <math>\frac{2017+1}{2}=1009</math> | ||
==See Also== | ==See Also== | ||
Revision as of 12:57, 9 May 2024
Problem 8
Let 
. The minimum value of 
is in the interval:
(a) ![$(-\infty,1008]$](http://latex.artofproblemsolving.com/9/c/3/9c3f0cd9ce7d4e89be5190f6dc60180d3363e96c.png)
(b) ![$(1008,1009]$](http://latex.artofproblemsolving.com/7/5/b/75bb7a0261dc6254465d88e70e4e960d0168dc6c.png)
(c) ![$(1009,1010]$](http://latex.artofproblemsolving.com/e/c/f/ecf6f844c56ebe845852158ed002a6a68904b550.png)
(d) ![$(1010,1011]$](http://latex.artofproblemsolving.com/0/c/8/0c8b0bdbefe87669231f4d789d7b60c2861cbd5d.png)
(e) 
Solution
The minimum value of this function is when 
is in the middle between 1 and 2017. 
