+2729 Let
f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor.
Evaluate f(1)+f(2) + f(3) + \dots + f(999)+f(1000). (This sum has 1000 terms, one for the result when we input each integer from 1 to 1000 into f.)
+410 The function inside, f(x)=⌊2−3x3x+1⌋ is 2−3x3x+1.
Using what we know about rational functions, set the inside function, to g(x), g(x)=2−3x3x+1 has a x intercept of 2/3, and a horizontal asymptote of -1, and a vertical asymptote at x = -1/3.
Using this information, we can determine the relative shape of the graph.
Because the function has an x intercept at 2/3, and the function only goes down after that, but will not go down past -1, since that is our horizontal asymptote. Plugging in values from 1≤x≤∞ will always result in a output of −1≤g(x)≤0.
Therefore because f(x)=⌊2−3x3x+1⌋, if x is from 1 to 1000 will always result in -1.
Therefore f(1)+f(2)+f(3)+⋯+f(999)+f(1000)=−1000.
+410 The function inside, f(x)=⌊2−3x3x+1⌋ is 2−3x3x+1.
Using what we know about rational functions, set the inside function, to g(x), g(x)=2−3x3x+1 has a x intercept of 2/3, and a horizontal asymptote of -1, and a vertical asymptote at x = -1/3.
Using this information, we can determine the relative shape of the graph.
Because the function has an x intercept at 2/3, and the function only goes down after that, but will not go down past -1, since that is our horizontal asymptote. Plugging in values from 1≤x≤∞ will always result in a output of −1≤g(x)≤0.
Therefore because f(x)=⌊2−3x3x+1⌋, if x is from 1 to 1000 will always result in -1.
Therefore f(1)+f(2)+f(3)+⋯+f(999)+f(1000)=−1000.
