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Modulus inequality

Find the range of x x values such that the inequality
∣∣2x−1∣−2∣≤(12)x \displaystyle \bigg ||2^x-1|-2 \bigg | \leq \left (\frac{1}{2}\right )^x
holds.

It's a bit of an awkward question.
(edited 7 years ago)
Original post by Ano123
Find the range of x x values such that the inequality
∣∣2x−1∣−2∣≤(12)x \displaystyle \bigg ||2^x-1|-2 \bigg | \leq \left (\frac{1}{2}\right )^x
holds.

It's a bit of an awkward question.


Pretty straight forward if you draw a graph. What about it?
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Ano123
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Original post by RDKGames
Pretty straight forward if you draw a graph. What about it?


What's your sketch?
Original post by Ano123
What's your sketch?


draw the graph
find the solution(s)
look at the range between/out of these solution(s) using the graph and bam u got yourself an answer :yy:
Original post by Ano123
What's your sketch?


Draw both sides as separate functions on the same axis
Original post by Ano123
What's your sketch?


Something like this.

ImageUploadedByStudent Room1473942846.250370.jpg


Posted from TSR Mobile
Original post by Ano123
Find the range of x x values such that the inequality
∣∣2x−1∣−2∣≤(12)x \displaystyle \bigg ||2^x-1|-2 \bigg | \leq \left (\frac{1}{2}\right )^x
holds.

It's a bit of an awkward question.


Consider the various cases. Firstly when x>0, then |2^x - 3| < 2^(-x) so if x>log2(3) then 2^x-3<2^(-x) and if 0<x<log2(3) then 3-2^x<2^(-x). Both of these are disguised quadratics, so you should be able to continue from here.

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