What is the value of M when |x|≤ 3 in the inequality (x^2+2x+1)/(x^2+3) ≤ M?

In summary: It turns out that there is a convenient way to solve such inequalities, which you can read about in my response to your other post.In summary, a value of M for the inequality |(x2+2x+1)/(x2+3)|≤ M can be found when |x|≤ 3. However, the statement that |u+v| = |u| + |v| is not always true. The absolute values in the inequality can be removed, and a quadratic inequality can be solved using a convenient method.
  • #1
altwiz
2
0
1. |(x2+2x+1)/(x2+3)|≤ M. Find the value of M when |x|≤ 3.



2. |u+v|=|u|+|v|



3. I understand that you start off by distributing the absolute value symbols into the individual terms as above. Then you maximize the numerator, using 3 as the value for x. However, my professor then minimized the denominator by using 0 as the value for x. This makes M = 240/3. What I do not understand is how you can use 2 different values of x to determine one value for M. Is there a value of x then that gives you M as the outcome in the L.H.S.?
 
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  • #2
altwiz said:
1. |(x2+2x+1)/(x2+3)|≤ M. Find the value of M when |x|≤ 3.



2. |u+v|=|u|+|v|



3. I understand that you start off by distributing the absolute value symbols into the individual terms as above. Then you maximize the numerator, using 3 as the value for x. However, my professor then minimized the denominator by using 0 as the value for x. This makes M = 240/3. What I do not understand is how you can use 2 different values of x to determine one value for M. Is there a value of x then that gives you M as the outcome in the L.H.S.?

Is x2 supposed to be x2? If so, either use the palette at the top of the input screen (as I just now did), or else write x^2.

Your statement |u+v| = |u| + |v| is false! For example, 0 = |1 + (-1)| ≠ |1| + |-1| = 2.
 
  • #3
As Ray already mentioned, your equation |u + v| = |u| + |v| is false.

The absolute values in your inequality can be removed, because x2 + 2x + 1 ≥ 0 for all real x, and x2 + 3 > 0 for all real x.
 
  • #4
I knew something was wrong with this question. Thank you for your replies, I will ask my professor for further clarification.
 
  • #5
altwiz said:
I knew something was wrong with this question. Thank you for your replies, I will ask my professor for further clarification.
Why? I don't think there's anything wrong with the question. Our complaint was with what you wrote as a relevant equation.

After doing some simplification, you wind up with a quadratic inequality.
 

Related to What is the value of M when |x|≤ 3 in the inequality (x^2+2x+1)/(x^2+3) ≤ M?

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