What am I supposed to do here?

blamocur said:
You are supposed to read the guildelines: https://www.freemathhelp.com/forum/threads/read-before-posting.109846/
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Don't post a list of homework problems.
it's not.
Try to post in an appropriate category.
this is from a calculus book, but even if I got it wrong, it says "try".
Try to use halfway-decent English.
pretty sure it's decent.
Post the complete text of the exercise.
it's the complete text
Show your beginning work, or ask a specific question about the exercise, or explain why you're stuck.
there is a line where I put the values of the domain in x places followed by a question mark since i don't know how to continue.
Proofread your posts for clarity.
i proofread it.
Be nice.
i am pretty sure what i wrote was not mean
Don't spam
i am pretty sure i didn't spam
Have patience.
I wasn't attacking anyone to hurry up.
Don't post URLs to image-hosting sites that add extra images to their pages.
didn't do that.
Don't create duplicate threads or posts.
also didn't do that.

anything i missed?
 
Loki123 said:
[-4,0] however i don't get how you got that....
Click to expand...
The function f has a domain of [0, 2]. So (x/2) + 2 has to be in that interval as well. So when you say that (x/2) + 2 is in [-4, 0]...

-Dan
 
You are given that 0<x<2.
Now get 2x in the middle of that inequality.
Get x/2 +2 in the middle of the given inequality.
Now what?
 
No, sorry but you are not getting it. You are given that 0<x<2. You can not change that fact. So it is not true that 0<2x<2. Try again.
 
I'll do the 1st one for you.
0<x<2. Now double all three terms. This gives you 0<2x<4
 
Loki123 said:
Don't post a list of homework problems.
it's not.
Try to post in an appropriate category.
this is from a calculus book, but even if I got it wrong, it says "try".
Try to use halfway-decent English.
pretty sure it's decent.
Post the complete text of the exercise.
it's the complete text
Show your beginning work, or ask a specific question about the exercise, or explain why you're stuck.
there is a line where I put the values of the domain in x places followed by a question mark since i don't know how to continue.
Proofread your posts for clarity.
i proofread it.
Be nice.
i am pretty sure what i wrote was not mean
Don't spam
i am pretty sure i didn't spam
Have patience.
I wasn't attacking anyone to hurry up.
Don't post URLs to image-hosting sites that add extra images to their pages.
didn't do that.
Don't create duplicate threads or posts.
also didn't do that.

anything i missed?
Click to expand...
I wish you would spend as much effort describing your attempts and thoughts as you spent on this response :) More seriously, the one line you refer to does not show, at least to me, what you've actually tried, where you are stuck, what kinds of questions you have, what kinds of mistakes you might be making, etc. I.e., it is difficult for me to see what is the best way to help you without actually solving the whole problem.
 
Loki123 said:
I get this. But how do I insert it into the entire expression? View attachment 32453
Click to expand...
I believe you have practically solved it here. You have two conditions for [imath]x[/imath], one for each part of the new expression. For the new function/expression to be defined both conditions must be met.
 
Steven G said:
Now think about how this will help you.
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my initial thought is to use 0 and 2 and calculate the expression for it - solution 1
then use 4 and 3 and calculate the expression - solution 2
and that the domain would be from solution 1 to solution 2.
 
You have to compute f(2x) and f(x/2 + 2). Not just one and then later on the other! You need to find x-values (domain) so you can compute both of them at the same time! This is after all what a domain is!
Post back.
 
Loki123 said:
my initial thought is to use 0 and 2 and calculate the expression for it - solution 1
then use 4 and 3 and calculate the expression - solution 2
and that the domain would be from solution 1 to solution 2.
Click to expand...
Maybe you do understand(?). What exactly do you mean by and that the domain would be from solution 1 to solution 2?
 
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