Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

Let f be a 1-1 function such that f(a) = b, f(b) = c, and f(c) = a. Find the following. f −1(f −1(c))

OpenStudy (anonymous):

deja vu

OpenStudy (anonymous):

i tryied asking you to explain more tryign to get it

OpenStudy (anonymous):

oh ok we take it one step at at time

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

we know \[f(a)=b\] so we also know that \[f^{-1}(b)=a\] so far ok?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

also we have \[f(b)=c\] so we know \[f^{-1}(c)=b\] right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so what does \[f^{-1}(f^{-1}(c))\] mean? it means first find \[f^{-1}(c)\] and then find \[f^{-1}\] of the result

OpenStudy (anonymous):

so writing it out we get \[f^{-1}(f^{-1}(c))=f^{-1}(b)\] and that is true because we know \[f^{-1}(c)=b\]

OpenStudy (anonymous):

next step is \[f^{-1}(b)=a\] which we also knew from above

OpenStudy (anonymous):

writing this in one line we get \[f^{-1}(f^{-1}(c))=f^{-1}(b)=a\] and that is what you wanted

OpenStudy (anonymous):

make sense or still confusing?

OpenStudy (anonymous):

okay i am getting it..so the answer is simple a or do i write if F-1(b)+a

OpenStudy (anonymous):

no answer is simply "a"

OpenStudy (anonymous):

there is no "+" involved

OpenStudy (anonymous):

meant = sorry

OpenStudy (anonymous):

sweet thank you i am pretty sure i get it now

OpenStudy (anonymous):

oh ok, the answer is just "a"

OpenStudy (anonymous):

for example if you know \[2x-1=7\] then you know that \[x=4\] this is like saying \[f(x)=2x-1\] and \[f(4)=7\] \[f^{-1}(7)=4\]

OpenStudy (anonymous):

okay cool thank you

OpenStudy (anonymous):

yw

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!