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OpenStudy (revircs):
OpenStudy (revircs):
Let k(x) = (f(x))^−1.
what is k'(2)?
OpenStudy (revircs):
I had the answer as -.12, but I dont remember how I got it.
OpenStudy (princeharryyy):
is it f(x) to the power -1
OpenStudy (revircs):
\[(f(x))^-1\] like this
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OpenStudy (revircs):
so yeah, (f(x)) to the power of -1
OpenStudy (princeharryyy):
ok
OpenStudy (princeharryyy):
what's that apostrophe on k(2) is it the differentiation.
OpenStudy (revircs):
yes
OpenStudy (princeharryyy):
are you sure, you got the right answer.
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OpenStudy (revircs):
Yeah, it marked it as correct for me
OpenStudy (princeharryyy):
See, from the information given you can not find f(x). Now, if that you can't do. You can not find the k(x). And, if it's not possible to find k(x) how can you find differentiation of k(x).
OpenStudy (revircs):
@baru any idea?
OpenStudy (princeharryyy):
from the information we only know that f(2) =5
and that if it is f(x) to the power -1
=>
k(2) = (5)^-1.
OpenStudy (revircs):
I found it out, we take the derivative of both sides.
so k'(x) = -f(x)^(-2) * (f'(x))