If #f(x^2) = 1/x# then what is #f(1/x)#?

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Answer 1 · 2017-10-07T20:36:32

Add the requirement #x > 0# to find:

#f(1/x) = sqrt(x)#

Given:

#f(x^2) = 1/x#

First note that we need an extra condition beyond what is specified in the question. Otherwise, putting #x=1# and #x=-1# we find:

#f(1) = f(1^2) = 1/1 = 1#

#f(1) = f((-1)^2) = 1/(-1) = -1#

So let us refine the given condition and specify:

#f(x^2) = 1/x" "# for any #x > 0#

We can replace the variable with any other of our choice.

So:

#f(t^2) = 1/t" "# for any #t > 0#

We want #t^2 = 1/x# and hence #t = +-1/sqrt(x)#

In order to use the given condition, we require #t > 0# so we need the positive square root here:

#t = 1/sqrt(x)#

Then:

#f(1/x) = f(t^2) = 1/t = 1/(1/sqrt(x)) = sqrt(x)#