What of the following statements have the same result?
f(1) when f(x) = 5x + 1
f−1(3) when f(x) = 2x+3
3y − 7 = y + 5
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
f(x) = 5x + 1
f(1) = 5(1) + 1 ... replace every x with 1
f(1) = ???
OpenStudy (anonymous):
5
jimthompson5910 (jim_thompson5910):
close but no
OpenStudy (anonymous):
6 lol
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
"5(1) + 1" means "5 times 1 plus 1"
jimthompson5910 (jim_thompson5910):
yeah f(1) = 6
jimthompson5910 (jim_thompson5910):
"f−1(3) when f(x) = 2x+3"
is a fancy way of saying "plug in f(x) = 3 and solve for x"
f(x) = 2x+3
3 = 2x+3
3-3 = 2x+3-3 ... Subtract 3 from both sides.
0 = 2x
x = ???
OpenStudy (anonymous):
0
jimthompson5910 (jim_thompson5910):
yep x = 0 leads to f(x) = 3
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
solve 3y − 7 = y + 5 for y. Tell me what you get
OpenStudy (anonymous):
6
jimthompson5910 (jim_thompson5910):
y = 6 is correct
jimthompson5910 (jim_thompson5910):
from part a) we got f(1) = 6
OpenStudy (anonymous):
ok so 1 and 3 :)
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
yeah
OpenStudy (anonymous):
I have one last one and I'm done for tonight!
jimthompson5910 (jim_thompson5910):
show me what you have so far
OpenStudy (anonymous):
Ok. so I guess a positive end behavior for f(x)+2
OpenStudy (anonymous):
yes?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
how do you know it's positive?
OpenStudy (anonymous):
i just assumbed bc the leading co. is positive
jimthompson5910 (jim_thompson5910):
we don't know anything about the original function f(x). So we can't say if it has positive or negative end behavior
jimthompson5910 (jim_thompson5910):
what we can say is that the end behavior won't change if we add 2 to f(x)
OpenStudy (anonymous):
true. I agree
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
f(x)+2 just shifts f(x) up 2 units
jimthompson5910 (jim_thompson5910):
on the other hand (-1/2)*f(x) flips f(x) over the x axis and compresses it vertically by a factor of 2
jimthompson5910 (jim_thompson5910):
so (-1/2)*f(x) will have its end behavior flipped
OpenStudy (anonymous):
I agree.
jimthompson5910 (jim_thompson5910):
what can you say about the y-intercept?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
It would be (0,2)
jimthompson5910 (jim_thompson5910):
but do we know what the y-intercept of f(x) is?
OpenStudy (anonymous):
and increasing?
OpenStudy (anonymous):
2
jimthompson5910 (jim_thompson5910):
we don't know what the y-intercept of f(x) is
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
but whatever it is, it is shifted up 2 units for f(x)+2
OpenStudy (anonymous):
yes. And what about the -1/2 one?
OpenStudy (anonymous):
so a wide parabola?
jimthompson5910 (jim_thompson5910):
it gets wider when you compress it vertically, yes
OpenStudy (anonymous):
But I don't understand how it is increasing and the regions where it does part>>
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
any ideas?
jimthompson5910 (jim_thompson5910):
f(x) isn't given, so we cannot find the increasing/decreasing intervals. Whatever they are, they don't change when going to f(x)+2. Everything shifts up which is why the intervals don't change
jimthompson5910 (jim_thompson5910):
with (-1/2)*f(x), the intervals swap. Whatever was decreasing is now increasing and vice versa.
OpenStudy (anonymous):
Oh:O !! That makes sooo much more sense! Thanks:)
OpenStudy (anonymous):
is that it? We are done with the question
Still Need Help?
Join the QuestionCove community and study together with friends!