ALGEBRA HELP .Last question
graph each function of g
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OpenStudy (aizhalee):
g(x)=f(2x)
OpenStudy (just_one_last_goodbye):
So coordinating points of G correct?
OpenStudy (aizhalee):
thank you for stopping by to help me :)
OpenStudy (aizhalee):
yes
OpenStudy (just_one_last_goodbye):
No prob xD
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OpenStudy (aizhalee):
thank you , so can you explain on how I can do that . Its a tad bit confusing
OpenStudy (just_one_last_goodbye):
are the coordinates already showing?
OpenStudy (just_one_last_goodbye):
aren't*
OpenStudy (aizhalee):
no using the graph that I posted up there to plug in . to solve the question
OpenStudy (just_one_last_goodbye):
@pooja195 please help my friend I dont understand the question
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OpenStudy (aizhalee):
its ok . I dont understand it too well either ;/
OpenStudy (just_one_last_goodbye):
stupid school questions @pooja195
OpenStudy (just_one_last_goodbye):
@SolomonZelman bro please help my friend ^_^ you very smart of these kind of questions
OpenStudy (just_one_last_goodbye):
@tkhunny please help
OpenStudy (tkhunny):
Domain of f(x) is what?
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OpenStudy (aizhalee):
-2, 2 @tkhunny
jimthompson5910 (jim_thompson5910):
from the graph, we see that f(2) = 2 because (2,2) is on the graph of f(x)
if we plug x = 2 into g(x), we get
g(x) = f(2x)
g(2) = f(2*2) .... replace each x with 2
g(2) = f(4)
g(2) = undefined
why undefined? Because the graph of f(x) doesn't stretch to x = 4. So f(4) is not defined
what happens when we plug in x = 1 into g(x) ?
OpenStudy (aizhalee):
Thanks a lot but if we plug in 1 . then 1*1= 1
OpenStudy (tkhunny):
So, "2x" has to be contained in [-2,2]. thus, Domain of g(x) is [-1,1]. So?
OpenStudy (aizhalee):
am I correct or did I do a mistake
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OpenStudy (aizhalee):
alright I'll just add these into my notes
jimthompson5910 (jim_thompson5910):
plug x = 1 into g(x), to get
g(x) = f(2x)
g(1) = f(2*1) .... replace each x with 1
g(1) = f(2)
g(1) = ???
OpenStudy (aizhalee):
defined
OpenStudy (aizhalee):
or undefined
jimthompson5910 (jim_thompson5910):
what is f(2) equal to?
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OpenStudy (aizhalee):
g(1)
jimthompson5910 (jim_thompson5910):
look at the graph
OpenStudy (aizhalee):
2
jimthompson5910 (jim_thompson5910):
yes, so hopefully you see how g(1) = 2 ?
OpenStudy (aizhalee):
Yes I agree
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jimthompson5910 (jim_thompson5910):
how about g(-1) ?
OpenStudy (aizhalee):
im guessing -2
jimthompson5910 (jim_thompson5910):
plug x = -1 into the g(x) function and use the f(x) graph
OpenStudy (aizhalee):
g(x) = f(2x)
g(-1) = f(2*-1) .... replace each x with 1
g(-1) = f(-2)
g(-1) = -2
jimthompson5910 (jim_thompson5910):
f(-2) is NOT equal to -2
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jimthompson5910 (jim_thompson5910):
draw a vertical line through x = -2 on the graph you have
where does this vertical line cross the f(x) graph? what is the y coordinate of this point?
OpenStudy (aizhalee):
4
OpenStudy (aizhalee):
if I do a vertical line there really isnt a y coordinate
OpenStudy (aizhalee):
i tried >.<
jimthompson5910 (jim_thompson5910):
you're looking at the graph where f(x) is a horizontal line through y = 2 right?
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OpenStudy (aizhalee):
yes
jimthompson5910 (jim_thompson5910):
you should see that x = -2 corresponds to y = 2 on the graph
so (-2,2) lies on f(x)
jimthompson5910 (jim_thompson5910):
agreed?
OpenStudy (aizhalee):
yes sir
jimthompson5910 (jim_thompson5910):
so hopefully you see how g(-1) = 2 as well?
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OpenStudy (aizhalee):
yes
OpenStudy (aizhalee):
whats next
jimthompson5910 (jim_thompson5910):
so as @tkhunny is saying, the domain of f(x) is [-2,2]
and the domain of g(x) shrinks to [-1,1]
so thinks have shrunk horizontally by a factor of 2
jimthompson5910 (jim_thompson5910):
the graph of g(x) is simply a horizontal line through y = 2 but it is restricted on the interval \(\Large -1 \le x \le 1\)
OpenStudy (aizhalee):
thanks a lot im adding this in my notes
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