Real parameter a

L

Loki123

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I think I got something wrong at a<=1. I don't think that's how you solve system i equations. What should I do? IMG_20220505_163001.jpg
 
Why do you exclude [imath]a=3/2[/imath] where [imath]x=1[/imath] seems a good solution ?
 
blamocur said:
True, but 3/2 is a legitimate value for [imath]a[/imath], which means that [imath]a \leq 1[/imath] is a wrong answer.
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i assumed it was, i don't know if it is possible to solve the system of inequalities like i did
but i am still left with an unanswered question
what statement is correct
i need some condition containing 1
 
i don't mean to be mean, but this is literally not helping me at all. your answers are too short and you just deny or confirm what i said without giving me any info.
 
Not to be mean, but my answers are supposed to be shorter than your questions. If I tell you that your particular statement is wrong please spend some effort and try explaining to me that it is correct. In my experience, making such effort often helps to discover errors. BTW, it took me a fraction of a second to see that not all answers mention 1. In fact, only 2 out 5 do -- why couldn't you do the same? You cannot expect me to point my finger at the exact place and tell you exactly what to do there -- I am in business of giving hints, not answers. I.e., it is much more important for me to help with understanding than with solving.

My other gripe is that you are often trying to figure out the answers to exersizes without making effort to understand the definitions and the concepts involved. E.g., if you are trying to solve a problem about continuous function you shouldn't wait for tutors to give you links to the corresponding definitions. A much more productive approach is to search for the term online (or in textbooks), try to understand it, then ask questions.
 
blamocur said:
Not to be mean, but my answers are supposed to be shorter than your questions. If I tell you that your particular statement is wrong please spend some effort and try explaining to me that it is correct. In my experience, making such effort often helps to discover errors. BTW, it took me a fraction of a second to see that not all answers mention 1. In fact, only 2 out 5 do -- why couldn't you do the same? You cannot expect me to point my finger at the exact place and tell you exactly what to do there -- I am in business of giving hints, not answers. I.e., it is much more important for me to help with understanding than with solving.

My other gripe is that you are often trying to figure out the answers to exersizes without making effort to understand the definitions and the concepts involved. E.g., if you are trying to solve a problem about continuous function you shouldn't wait for tutors to give you links to the corresponding definitions. A much more productive approach is to search for the term online (or in textbooks), try to understand it, then ask questions.
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you can't blame me for not understanding hints
i am not looking for answers but a bit more advice on how to solve the problem rather than pointing out that it's wrong.
i know it's wrong that's why i am here and i also wouldn't be here if i knew where i went wrong
my question in the op was, what should i do
i am trying to understand but i need a bit more
my school system does not cover all this information
english is not my first language so i don't even know the correct terms
 
I never blame people for not understanding, nor would I blame anyone for making mistakes. But when we disagree it is more important for you to explain your statements then for me to explain mine. In my pretty long experience trying to explain something often led me to uncover my own errors and discover holes in my own understanding of math concepts.

In the past I spent a lot of time trying, often in vain, to understand various parts of math purely from books and without the benefit of help. I know how frustrating this can be, and I admire your perseverance. But you will be more successful if you pay more attention to the concepts. E.g., "I wonder what does it actually mean for a function to be continuous at a given point".

BTW., I have no problem with your English, maybe because it's not my first language either. Also, I believe math is the least language-dependent part of science.
 
But when we disagree it is more important for you to explain your statements then for me to explain mine.
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Forgot to mention another reason: when you give me your explanations it might help me see what you are missing or misunderstanding, which, in turn, can help me give you better hints. And occasionally your explanations can show me that I am wrong, which happens too.
 
blamocur said:
Also, I believe math is the least language-dependent part of science.
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I respectfully disagree, especially coming from a background in Physics. Math terms are a dictionary and without the precise definitions it means essentially nothing. I suspect that is part of what is happening here. I have seen Loki123 getting lost simply because (s)he needed to have the definition spelled out. Language problem? Possibly. But looking up the definition should theoretically help that alot.

-Dan
 
topsquark said:
I respectfully disagree, especially coming from a background in Physics. Math terms are a dictionary and without the precise definitions it means essentially nothing. I suspect that is part of what is happening here. I have seen Loki123 getting lost simply because (s)he needed to have the definition spelled out. Language problem? Possibly. But looking up the definition should theoretically help that alot.

-Dan
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How do physicists say "[imath]\forall \epsilon>0 \exists\delta : |x-a|\leq \delta \rightarrow |f(x)-b|\leq\epsilon[/imath]" ?;)
 
Excuse me for intruding late into a long thread.

I am a great believer that it would help students if we stress in algebra that learning mathematics requires learning two types of language. One type has just one exemplar, namely the formal notation of mathematics. That is an international written language, and in its pure form involves no sounds at all. The second type is the technical vocabulary used to discuss mathematics and to translate mathematical notation into a natural spoken language. Each natural language has its own technical vocabulary for mathematics. I have no idea how to write or say the Japanese version of "inverse operation."

"f(x) is continuous at a" in English normally has the following meaning:

[math]f(a) \in \mathbb R \ \land f(a) = \lim_{x \rightarrow a}f(x)[/math]
It would help if we explicitly recognized the two types of language involved in mathematics.
 
JeffM said:
Excuse me for intruding late into a long thread.

I am a great believer that it would help students if we stress in algebra that learning mathematics requires learning two types of language. One type has just one exemplar, namely the formal notation of mathematics. That is an international written language, and in its pure form involves no sounds at all. The second type is the technical vocabulary used to discuss mathematics and to translate mathematical notation into a natural spoken language. Each natural language has its own technical vocabulary for mathematics. I have no idea how to write or say the Japanese version of "inverse operation."

"f(x) is continuous at a" in English normally has the following meaning:

[math]f(a) \in \mathbb R \ \land f(a) = \lim_{x \rightarrow a}f(x)[/math]
It would help if we explicitly recognized the two types of language involved in mathematics.
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Excuse me, but this definition is not quite correct. It works only if a is limit point of domain of the function [imath]f(x)[/imath], because limit of a function defined only for such points. For readers, the correct definition is following:

Let [imath]f:E \rightarrow \mathbb{R}, where E \subset \mathbb{R}[/imath] and [imath]a \in E[/imath]. [imath]f(x)[/imath] is continious at point [imath]a[/imath] if [imath]\forall \varepsilon > 0 \exists \delta>0: |x-a|<\delta \Rightarrow |f(x)-f(a)| < \varepsilon[/imath].
 
Danteth said:
Excuse me, but this definition is not quite correct. It works only if a is limit point of domain of the function [imath]f(x)[/imath], because limit of a function defined only for such points. For readers, the correct definition is following:

Let [imath]f:E \rightarrow \mathbb{R}, where E \subset \mathbb{R}[/imath] and [imath]a \in E[/imath]. [imath]f(x)[/imath] is continious at point [imath]a[/imath] if [imath]\forall \varepsilon > 0 \exists \delta>0: |x-a|<\delta \Rightarrow |f(x)-f(a)| < \varepsilon[/imath].
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I think that your point is captured in the definition of limit.
 
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