Don't post a list of homework problems.You are supposed to read the guildelines: https://www.freemathhelp.com/forum/threads/read-before-posting.109846/
[-4,0] however i don't get how you got that....What x values are there such that (x/2) + 2 is in [0, 2]?
-Dan
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]...[-4,0] however i don't get how you got that....
That's the domain of that functionThe 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
but that's how you worded it?? i am confusedNo, 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.
Now get 2x in the middle of that inequality.
Get x/2 +2 in the middle of the given inequality.
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.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?
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.I get this. But how do I insert it into the entire expression? View attachment 32453
got itI'll do the 1st one for you.
0<x<2. Now double all three terms. This gives you 0<2x<4
Now think about how this will help you.got it
for 2x it's [0,4]
for x/2+2 it's [2,3]
now what?
my initial thought is to use 0 and 2 and calculate the expression for it - solution 1Now think about how this will help you.
Maybe you do understand(?). What exactly do you mean by and that the domain would be from solution 1 to solution 2?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.
solution 1 = -1Maybe you do understand(?). What exactly do you mean by and that the domain would be from solution 1 to solution 2?