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

[altwiz]

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.?

 

[Ray Vickson]

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.

 

[Mark44]

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.

 

[altwiz]

I knew something was wrong with this question. Thank you for your replies, I will ask my professor for further clarification.

 

[Mark44]

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.