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I'm told this a very easy problem once you figure it out. Unfortunately, I cannot seem to get past the first step.

The Identity Function I Behaves Just Like 1.

That is, f o I = f And I o g = g

It is also true that for the function K, K o K^-1 = I

Use this information to solve for f, h = f o g

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  • $\begingroup$ Is there enough information to provide an answer? $\endgroup$
    – mlc
    Commented May 2, 2017 at 6:12
  • $\begingroup$ There should be. Like I said, this is supposed to be a relatively easy problem, I'm just missing something obvious. $\endgroup$
    – Michael McDon
    Commented May 2, 2017 at 6:14
  • $\begingroup$ What kind of functions are you referring to in this question? $\endgroup$
    – Lazy Lee
    Commented May 2, 2017 at 6:16
  • $\begingroup$ Wouldn't it be convenient if you could cancel out that $g$ in $h = f \circ g$? How would you manage that? $\endgroup$
    – Brian Tung
    Commented May 2, 2017 at 6:18

1 Answer 1

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Assuming $g^{-1}$ exists.

$$h = f \circ g$$

$$h \circ g^{-1} = f \circ g \circ g^{-1}$$

$$f = h \circ g^{-1}$$

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