MHB Real Solutions for Equation |x-|x-|x-4||| = a

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The discussion centers on finding real numbers \( a \) for which the equation \( |x-|x-|x-4||| = a \) yields exactly three real solutions. Participants explore the behavior of the nested absolute value function and its critical points to determine the values of \( a \). The conversation highlights the importance of understanding the structure of the equation to identify when it intersects with horizontal lines representing \( a \). Additionally, there is acknowledgment of a member's successful method in solving the equation, indicating a shared approach among participants. The focus remains on the mathematical analysis required to solve the equation effectively.
anemone
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Find all real numbers $$a$$ such that the equation $$ |x-|x-|x-4||| =a
$$ has exactly three real solutions.
 
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anemone said:
Find all real numbers $$a$$ such that the equation $$ |x-|x-|x-4||| =a
$$ has exactly three real solutions.


The curve

$$y= |x-|x-|x-4||| $$

is piece-wise linear with knots at $$x=4/3,\ 2,\ 4$$ and so considering the shape of the curve $$a=y(2)$$ is the value sought.
.
 
zzephod said:
The curve

$$y= |x-|x-|x-4||| $$

is piece-wise linear with knots at $$x=4/3,\ 2,\ 4$$ and so considering the shape of the curve $$a=y(2)$$ is the value sought.
.

The OP is having trouble logging on to MHB, and has asked me to congratulate you on her behalf on finding the correct answer, and says the method you have used is nearly identical to the method she used in finding the solution.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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