Proving g(x) is continuous over interval (-∞,-2)

  • Thread starter Thread starter member 731016
  • Start date Start date
  • Tags Tags
    Continuous Interval
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For number 18,
1676597607436.png

The solution is,
1676597636101.png

However, should they not write "For ## -∞ < a < -2##" since ##a ≠ -∞## (infinity is not a number)?

Many thanks!
 
Physics news on Phys.org
##a## is a number, so can't be negative infinity.

In particular, ##a\in [-\infty,-2)## isn't really a meaningful thing outside of people abusing notation.
 
  • Like
Likes member 731016
Office_Shredder said:
##a## is a number, so can't be negative infinity.

In particular, ##a\in [-\infty,-2)## isn't really a meaningful thing outside of people abusing notation.
Thank you for your reply @Office_Shredder !

So a better notation than ## a < -2 ## is ## -∞ < a < -2##, correct?

Many thanks!
 
Callumnc1 said:
Thank you for your reply @Office_Shredder !

So a better notation than ## a < -2 ## is ## -∞ < a < -2##, correct?

Many thanks!
No, the ##-\infty<a## notation conveys no additional information. ##a## and ##a>-\infty## are the same thing. You can include the negative infinity, but in no sense is it better here.
 
  • Like
Likes member 731016
Callumnc1 said:
However, should they not write "For −∞<a<−2"
##-\infty < a < -2## and ##a \in (-\infty, -2)## are two different notations that say exactly the same thing. In most of the books I've seen, interval notation, as in the 2nd example above, uses a parenthesis to indicate an endpoint that isn't included, and a bracket to indicate that an endpoint is included. I've seen other notations used, but these seem to be a lot rarer.

Also, I don't think any textbook would include "For" in a compound inequality. ##-\infty < a < -2## says everything that needs to be said.
 
  • Like
Likes member 731016
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top