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Mathematics 15 Online
OpenStudy (kaitlyn_nicole):

Let f(x) = x^2 − 81. Find f−1(x).

OpenStudy (anonymous):

this one is different

OpenStudy (kaitlyn_nicole):

yeah, i thought so

OpenStudy (anonymous):

first off, it is not a one to one function, so it has no inverse, but never mind that, go ahead and solve \[x=y^2-81\] for \(y\)

OpenStudy (kaitlyn_nicole):

how do i solve with two variables?

OpenStudy (anonymous):

the same way you would solve \[19=x^2-81\]

OpenStudy (kaitlyn_nicole):

o how'd you get 19? or are you just putting in a random number

OpenStudy (anonymous):

add \(81\), then take the square root of both sides, although your answer will have an \(x\) in it

OpenStudy (anonymous):

i just picked a random number as an example

OpenStudy (anonymous):

the procedure would be the same for any number, and therefore any variable add 81 take the square root

OpenStudy (kaitlyn_nicole):

ok, so x= y^2-81

OpenStudy (anonymous):

ok now add 81

OpenStudy (kaitlyn_nicole):

x-81=y^2

OpenStudy (anonymous):

hmm no

OpenStudy (kaitlyn_nicole):

i mean plus

OpenStudy (anonymous):

add, not subtract

OpenStudy (anonymous):

yeah that

OpenStudy (kaitlyn_nicole):

and then x+9=y?

OpenStudy (anonymous):

oh no!

OpenStudy (kaitlyn_nicole):

oo wat

OpenStudy (anonymous):

just write a square root sign over it

OpenStudy (anonymous):

\[x+81=y^2\\ \pm\sqrt{x+81}=y\]

OpenStudy (kaitlyn_nicole):

ooooh snap

OpenStudy (kaitlyn_nicole):

you're really good at this lool

OpenStudy (anonymous):

now that \(\pm\) out front means you do not have a function, so the question was in error

OpenStudy (kaitlyn_nicole):

oh okay, i was wondering what that meant.

OpenStudy (anonymous):

for \[f(x)=x^2-81\] two different inputs will give the same output

OpenStudy (anonymous):

for example \[f(9)=81-81=0\] and \[f(-9)=0\] too so it does not have an inverse

OpenStudy (kaitlyn_nicole):

and they both fall on 0 so its not a function

OpenStudy (anonymous):

that is right since \((9,0)\) and \((-9,0)\) are on the graph of \(f\), the graph of \(f^{-1}\) would have \((0,9)\) and also \((0,-9)\) one input with two different outputs, not a function

OpenStudy (anonymous):

where did this question come from, FLVS or sommat?

OpenStudy (kaitlyn_nicole):

flvs

OpenStudy (kaitlyn_nicole):

alg 2.

OpenStudy (anonymous):

BINGO!!

OpenStudy (anonymous):

got it on the first try they really need to get their math act together down in Fla

OpenStudy (kaitlyn_nicole):

bingoooo

OpenStudy (kaitlyn_nicole):

yeah im really bad at math looool

OpenStudy (anonymous):

no it is not your fault at all

OpenStudy (anonymous):

you are asked bad questions, what can you do? don't they have any real schools near you?

OpenStudy (danjs):

i always think of the inverse switching the domain and range sets around on the graph the two look mirrored over the line y=x, since all x and y swap to the other reverses the function

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