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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.)

 Mar 9, 2024

Best Answer 

 #1
avatar+394 
+3

The function inside, \(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor\) is \(\frac{2-3x}{3x+1}\).

Using what we know about rational functions, set the inside function,  to g(x), \(g(x) = \frac{2-3x}{3x+1}\) 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 \(1\le x \le \infty\) will always result in a output of \(-1\le g(x) \le 0\).

Therefore because \(f(x)=\lfloor\frac{2-3x}{3x+1}\rfloor\), if x is from 1 to 1000 will always result in -1.

Therefore \(f(1)+f(2)+f(3) + \dots + f(999) + f(1000) = -1000\).

 Mar 9, 2024
 #1
avatar+394 
+3
Best Answer

The function inside, \(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor\) is \(\frac{2-3x}{3x+1}\).

Using what we know about rational functions, set the inside function,  to g(x), \(g(x) = \frac{2-3x}{3x+1}\) 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 \(1\le x \le \infty\) will always result in a output of \(-1\le g(x) \le 0\).

Therefore because \(f(x)=\lfloor\frac{2-3x}{3x+1}\rfloor\), if x is from 1 to 1000 will always result in -1.

Therefore \(f(1)+f(2)+f(3) + \dots + f(999) + f(1000) = -1000\).

hairyberry Mar 9, 2024

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