This problem is equivalent to a variant of a problem.
Let f (x) = | x+1|+| x+2 |+... | x+100 |,
(1) When x≤- 100,
f(x)=-[(x+ 1)+(x+2)+……+(x+ 100)]
=- 100 x-5050≥4950 & gt； 2009,
(2) When x≥- 1,
f(x)=(x+ 1)+(x+2)+……+(x+ 100)
= 100 x+5050≥4950 >； 2009,
(3) When-100
Then the minimum value of f(x) is
f(-50)
=[(50- 1)+(50-2)+……+(50-50)]+[(5 1-50)+(52-50)+……+( 100-50)]=2500.
In fact, when -5 1≤x≤-50, it is always equal to the minimum value f (x) = 2500.
Comprehensive (1)(2)(3) available: with f (x) applicable to all x >: 2009.
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To answer the landlord's question, first of all, we know that |a-b| geometrically represents the distance between two points A and B on the number axis.
So f (x) = | x+1|+| x+2 |+...+| x+100 |
It means that the electricity M with coordinate X reaches the point A 1, -2, -3, ..., -99,-100, A2, A3, ..., A99, A 100.
Sum of distances | ma1|+ma2 |+| ma3 |+...+| ma100 |.
(1-00) When x is less than or equal to-100, m falls in the position of m'.
f(x)= | x+ 1 |+| x+2 |+……+| x+ 100 |
= | ma 1 |+| MA2 |+| MA3 |+……+| ma 100 |
& gt| a 100 a 1 |+| a 100 a2 |+| a 100 a3 |+……+| a 100 a99 |+| a 100 a 100 |
=99+98+97+……+ 1+0=4950.
From the point of view of function increase and decrease
f(x)= | x+ 1 |+| x+2 |+……+| x+ 100 |
=-(x+ 1)-(x+2)-(x+3)-(x+ 100)
=- 100x-5050，
Monotonically decreasing, the minimum value is f(- 100)=4950.
(2-00) When x≥- 1, m falls on the position of m ",
f(x)= | x+ 1 |+| x+2 |+……+| x+ 100 |
= | ma 1 |+| MA2 |+| MA3 |+……+| ma 100 |
& gt| a 1a 1 |+| a 1 a2 |+| a 1 a3 |+……+| a 1a 99 |+| a 1a 100 |
=0+ 1+2+3+……+98+99=4950,
From the point of view of function increase and decrease
f(x)= | x+ 1 |+| x+2 |+……+| x+ 100 |
=(x+ 1)+(x+2)+(x+3)+……+(x+ 100)
= 100x+5050，
Monotonically increasing, the minimum value is f(- 1)=4950.
(1-0 1) When-100≤x≤-99,
f(x)= | x+ 1 |+| x+2 |+……+| x+98 |+| x+99 |+| x+ 100 |
=[| x+ 1 |+| x+2 |+……+| x+98 |]+[| x+99 |+| x+ 100 |]
=[| x+ 1 |+| x+2 |+……+| x+98 |]+[-(x+99)+(x+ 100)]
= [-(x+1)-(x+2)-...-(x+98)]+1monotonically decreasing,
The maximum value is f(- 100)=4950,
The minimum value is f(-99)=485 1.
(2-0 1) When -2≤x≤- 1,
f(x)= | x+ 1 |+| x+2 |+| x+3 |+……+| x+ 100 |
=[| x+ 1 |+| x+2 |]+[| x+3 |+……+| x+ 100 |]
=[-(x+)+(x+2)]+[| x+3 |+……+| x+ 100 |]
=1+[(x+3)+...+(x+100)] monotonically increases,
The maximum value is f(- 1)=4950,
The minimum value is f(-2)=4852.
(1-02) when -99
f(x)= | x+ 1 |+| x+2 |+……+| x+96 |+| x+97 |+| x+98 |+| x+99 |+| x+ 100 |
=[| x+ 1 |+| x+2 |+……+| x+96 |]+[| x+97 |+| x+98 |+| x+99 |+| x+ 100 |]
=[| x+ 1 |+| x+2 |+……+| x+96 |]+[-(x+97)-(x+98)+(x+99)+(x+ 100)]
=[| x+ 1 |+| x+2 |+……+| x+96 |]+4
= [-(x+1)-(x+2)-...-(x+96)]+4 monotonically decreases,
The maximum value is f(-99)=4852,
The minimum value is f(-98)=4756.
(2-02) When -3≤x≤-2,
f(x)=[| x+ 1 |+| x+2 |+| x+3 |+| x+4 |]+[| x+5 |+……+| x+ 100 |]
=[-(x+ 1)-(x+2)+(x+3)+(x+4)]+[| x+5 |+……+| x+ 100 |]
= 4+[(x+5)+...+(x+ 100)] monotonically increases,
The maximum value is f(-2)=4852,
The minimum value is f(-3)=4756.
…… ……
(1-49) When -52
f(x)= | x+ 1 |+| x+2 |+| x+3 |+| x+4 |+| x+5 |+……+| x+98 |+| x+99 |+| x+ 100 |
= | x+ 1 |+| x+2 |+[| x+3 |+| x+4 |+| x+5 |+……+| x+98 |+| x+99 |+| x+ 100 |]
=-(x+ 1)-(x+2)+[-(x+3)-(x+4)-……+(x+98)+(x+99)+(x+ 100)]
=-(x+1)-(x+2)+2401=-2x+2398 monotonically decreasing,
The maximum value is f(-52)=2502,
The minimum value is f(-5 1)=2500.
(2-49) When -50≤x≤-49,
f(x)= | x+ 1 |+| x+2 |+| x+3 |+……+| x+4 |+| x+5 |+| x+98 |+| x+99 |+| x+ 100 |
=[| x+ 1 |+| x+2 |+| x+3 |+……+| x+96 |+| x+97 |+| x+98 |]+| x+99 |+| x+ 100 |
=[-(x+ 1)-(x+2)-(x+3)-(x+96)+(x+97)+(x+98)]+(x+99)+(x+ 100)
=2600+2x, monotonically increasing,
The maximum value is f(-49)=2502,
The minimum value is f(-50)=2500.
(1-50) When -5 1≤x≤-50,
(2-50) When -5 1≤x≤-50,
f(x)= = | x+ 1 |+| x+2 |+| x+3 |+……+| x+50 |+| x+5 1 |+……+| x+98 |]+| x+99 |+| x+ 100 |
= | x+ 1 |+| x+2 |+| x+3 |+……+| x+50 |+| x+5 1 |+……+| x+98 |]+| x+99 |+| x+ 100 |
=-(x+ 1)-(x+2)-(x+3)-(x+50)+(x+5 1)+……+(x+98)]+(x+99)+(x+ 100)
=2500, which is a constant.