Find all values such that the inequality is true

  • Thread starter JOATMON
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  • #1
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Mod note: Moved from a technical math section, so missing the homework template.
This is for an Intro to Analysis course. It's been a very long time since I've taken a math course, so I do not remember much of anything.

=============
Here is the problem:


For the inequality below, find all values n ∈ N such that the inequality is true:

(n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025.


============================
Here is my attempt at the problem:


Looking at the following set

{n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


We want to find the lower bound of this set.


Suppose A denotes the above set, then we have


A= {n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


Since the above rational function can be reduced to 1/(2n+1) we have


1/(2n+1) <0.025


Where we get n>19.5. Since the lower bounds of the set are 19.5, 19.4, 19.3.... And so on, this set is bound below by the natural number n=19. Therefore, the above inequality holds true for all n ∈ N greater than or equal to 20.


==========
If you can only just tell me what topics I need to review to answer this correctly, I would appreciate it.
 
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Answers and Replies

  • #2
ChrisVer
Gold Member
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what is n2,n3?
for example if n=1, n2=1 and n3=170000 you can make the inequality true....
[itex]\frac{1+2+3}{2 \times 170000 + 5 + 8 +3} = \frac{6}{340016}= 0.00001764622 < 0.025[/itex]
and so n=1 is keeping the inequality
 
  • #3
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what is n2,n3?

Sorry. I didn't check to see if the format changed once I copied and pasted my problem:
==============================================================

Here is the problem:


For the inequality below, find all values n ∈ N such that the inequality is true:

(n2 + 2n +3) / (2n3 + 5n2+ 8n + 3) < 0.025.


============================
Here is my attempt at the problem:


Looking at the following set

{n ∈ N: (n2+ 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


We want to find the lower bound of this set.


Suppose A denotes the above set, then we have


A= {n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


Since the above rational function can be reduced to 1/(2n+1) we have


1/(2n+1) <0.025


Where we get n>19.5. Since the lower bounds of the set are 19.5, 19.4, 19.3.... And so on, this set is bound below by the natural number n=19. Therefore, the above inequality holds true for all n ∈ N greater than or equal to 20. Thus A= {20, 21, 22, 23,...}
 
  • #4
35,499
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I don't see the purpose of all those "steps" before reducing the rational function (which you should probably show in more detail).
Where we get n>19.5
Okay (showing the steps wouldn't hurt).
Since the lower bounds of the set are 19.5, 19.4, 19.3.... And so on
What is the relevance of those numbers?

Once you have n>19.5 you can directly conclude that n>=20, and write down the set of solutions.
 
  • #5
RUber
Homework Helper
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One method that might help with the simplification is to flip the inequality.
##\frac{ n^2 +2n+3}{2n^3 + 5n^2 + 8n+3} < .025 \equiv \frac{2n^3 + 5n^2 + 8n+3}{ n^2 +2n+3}>40##
Which as you pointed out can be written as
##2n+1 > 40.##
And you already have the solution.
 
  • #6
35,499
11,946
Be careful: to do that you have to check that the two sides cannot be negative. This is easy to do here, but it is a necessary step.
 

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