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1. The greatest integer function, \(\lfloor x\rfloor\), denotes the largest integer less than or equal to x. For example, \(\lfloor3.5\rfloor=3\), \(\lfloor\pi\rfloor=3\) and \(\lfloor -\pi\rfloor=-4\). Find the sum of the three smallest positive solutions to \(x-\lfloor x\rfloor=\frac1{\lfloor x\rfloor}.\) Express your answer as a mixed number.

 

2. Let g(x) be a function piecewise defined as

\(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\)

If a is negative, find a so that \(g(g(g(10.5)))=g(g(g(a))).\)

 

Thank you 

 Sep 16, 2019
 #1
avatar+23850 
+1

1.
The greatest integer function, \(\lfloor x\rfloor\)r, denotes the largest integer less than or equal to \(x\). For example, \(\lfloor3.5\rfloor=3\), and \(\lfloor -\pi\rfloor=-4\).

 

Find the sum of the three smallest positive solutions to \(x-\lfloor x\rfloor=\frac1{\lfloor x\rfloor}\).
Express your answer as a mixed number.

 

\(\text{Let $ \text{frac}(x) = x - \lfloor x\rfloor $ } \\ \text{Let $ x = \lfloor x\rfloor + \text{frac}(x) $ } \)

 

I assume:

\(\begin{array}{|rcll|} \hline x-\lfloor x\rfloor &=& \frac1{\lfloor x\rfloor} \quad | \quad x - \lfloor x\rfloor = \text{frac}(x) \\ \text{frac}(x) &=& \frac1{\lfloor x\rfloor} \\ \mathbf{\lfloor x\rfloor \cdot \text{frac}(x)} &=& \mathbf{1} \\ \hline \end{array} \)

 

\(\begin{array}{|rclcl|c|} \hline \mathbf{\lfloor x\rfloor } &\cdot& \mathbf{\text{frac}(x)} &=& \mathbf{1}& x = \lfloor x\rfloor + \text{frac}(x) \\ \hline 2 &\cdot& \dfrac{1}{2} &=& 1 & 2\dfrac{1}{2} \\ \hline 3 &\cdot& \dfrac{1}{3} &=& 1 & 3\dfrac{1}{3} \\ \hline 4 &\cdot& \dfrac{1}{4} &=& 1 & 4\dfrac{1}{4} \\ \hline 5 &\cdot& \dfrac{1}{5} &=& 1 & 5\dfrac{1}{5} \\ \hline 6 &\cdot& \dfrac{1}{6} &=& 1 & 6\dfrac{1}{6} \\ \hline &\ldots &&&&\ldots \\ \hline \end{array} \)

 

The sum of the three smallest positive solutions is \(2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4} = 10\dfrac{1}{12} = 10.08\overline{3}\)

 

laugh

 Sep 17, 2019
 #2
avatar
+1

THX :)

 Sep 17, 2019

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