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# Ridiculous Inverse

## Question

Many of my algebra and precalculus students think the 'inverse function' of f(x), often written f^(-1)(x), is the same as the reciprocal 1/f(x) (mistaking the -1 for an exponent). This (as I am obliged to remind them) is almost always false. But can you find at least one function whose inverse is also its reciprocal? Tiebreaker: Find as many as you can!
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## 5 answers to this question

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For starters,

f^-1(f(x)) = x

but you told us that:
f^-1(f(x)) = 1/f(x)

putting these together,
f(x) = 1/x

Other than tricky ones, to me seems to be the only such function... Thinking of tricky ones now ...
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There are a few functions that are their own inverses. 1/x (away from 0) is one instance.

For tiebreaker are you asking for other instances of functions for which f[f(x)]= x?

Or for more functions whose inverses are their reciprocals?

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There are a few functions that are their own inverses. 1/x (away from 0) is one instance.

For tiebreaker are you asking for other instances of functions for which f[f(x)]= x?

Or for more functions whose inverses are their reciprocals?

The latter

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this obviously only applies to invertible functions and since invertible functions have exactly one inverse, I claim 1/x is the only such function.

For starters,

f^-1(f(x)) = x

but you told us that:

f^-1(f(x)) = 1/f(x)

putting these together,

f(x) = 1/x

Other than tricky ones, to me seems to be the only such function... Thinking of tricky ones now ...

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this obviously only applies to invertible functions and since invertible functions have exactly one inverse, I claim 1/x is the only such function.

For starters,

f^-1(f(x)) = x

but you told us that:

f^-1(f(x)) = 1/f(x)

putting these together,

f(x) = 1/x

Other than tricky ones, to me seems to be the only such function... Thinking of tricky ones now ...

I think the OP means that f^-1(x)=1/f(x).

One group fo solutions would be ones of the form (ax+b)/(cx+d) for appropriate a,b,c,d.

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