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zmudz

  • one year ago

Find all \(x\) for which \(\left| x - \left| x-1 \right| \right| = \lfloor x \rfloor\) The answer is suppose to be in a range, but when I put in that it can be all real numbers, that is wrong.

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  1. dinamix
    • one year ago
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    x=1 and x= 1/3

  2. anonymous
    • one year ago
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    I would try using the definition of the absolute value function to rewrite the LHS into something that you can easily match up with the RHS. \[|x|=\begin{cases}x&\text{for }x>0\\0&\text{for }x=0\\-x&\text{for }x<0\end{cases}\] This means \[\begin{align*}|x-|x-1||&=\begin{cases}x-|x-1|&\text{for }x-|x-1|>0\\0&\text{for }x-|x-1|=0\\-x+|x-1|&\text{for }x-|x-1|<0\end{cases}\\[2ex] &=\begin{cases}x-|x-1|&\text{for }x>\dfrac{1}{2}\\[1ex]0&\text{for }x=\dfrac{1}{2}\\[1ex]-x+|x-1|&\text{for }x<\dfrac{1}{2}\end{cases} \end{align*}\] You also have \[\begin{align*} |x-1|&=\begin{cases}x-1&\text{for }x>1\\0&\text{for }x=1\\-x+1&\text{for }x<1\end{cases} \end{align*}\]

  3. anonymous
    • one year ago
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    Putting all this together, you get \[\begin{align*} |x-|x-1||&= \begin{cases} 1&\text{for }x>1\\x&\text{for }x=1\\[1ex]2x-1&\text{for }\dfrac{1}{2}<x<1\\[1ex]0&\text{for }x=\dfrac{1}{2}\\[1ex]-2x+1&\text{for }x<\dfrac{1}{2} \end{cases} \end{align*}\]

  4. anonymous
    • one year ago
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    Now, given that \[\lfloor x\rfloor=\begin{cases}&\vdots\\ 0&\text{for }0\le x<1\\ 1&\text{for }1\le x<2 \\&\vdots\end{cases}\] you have the isolated solution \(x=\dfrac{1}{2}\) (since both LHS and RHS are zero) and the range \([1,2)\).

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