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The function $f$ is given as $f(x)=ax-1$ and the function $g$ as $g(x)=(x+1)/3$.

If $(f \circ g)(x)$ is an identity function, what is the value of $a$?

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Sep 26 '18 at 9:31
  • $\begingroup$ I think $f(g(x)) = f((x+1)/3) = (ax+a)/3+1$, not $-1$, right ?? $\endgroup$ – Anik Bhowmick Sep 26 '18 at 9:36
  • $\begingroup$ I think the question should be $f(x) = ax-1$. It was this way originally in the body. The title has always wrong. I believe the body has now been incorrectly edited to be consistent with the (incorrect) title. $\endgroup$ – Deepak Sep 26 '18 at 10:00
  • $\begingroup$ Ah good, that's been fixed. $\endgroup$ – Deepak Sep 26 '18 at 10:02
  • $\begingroup$ Both the printer and the OP may want to check signs. $\endgroup$ – Oscar Lanzi Sep 26 '18 at 10:02
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There was a misunderstanding.

Edit: $$fog(x)= a\dfrac{x+1}{3}-1$$ Let,$$y(x)=fog(x)$$ So,now,for to be identity function, $$y(x)=x$$ $$fog(x)=x$$ $$\implies a\dfrac{x+1}{3}-1= x$$ $$a=3$$

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  • $\begingroup$ The identity function is $x$. $\endgroup$ – Deepak Sep 26 '18 at 11:39
  • $\begingroup$ $f(x)=x \implies f(g(x))=g(x)$ $\endgroup$ – Rakibul Islam Prince Sep 26 '18 at 11:44
  • $\begingroup$ No, the composition $fg$ is the identity. So $fg(x) = x$. $\endgroup$ – Deepak Sep 26 '18 at 11:45
  • $\begingroup$ Yes,I agree, the composite function is identity.,so $fog(x)=f(g(x))$ has to be identity. that is what i have written. $\endgroup$ – Rakibul Islam Prince Sep 26 '18 at 11:49
  • $\begingroup$ How is this what you have written? You've written $fg(x) = g(x)$, whereas what you should have written is $fg(x) = x$. $\endgroup$ – Deepak Sep 26 '18 at 11:52
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I'm unsure about what your question is supposed to be, because I remember seeing a minus sign in there before the latest edit, i.e. $f(x) = ax-1$.

If your question is indeed that (i.e. $f(x) = ax-1$) with all else being unchanged, then it has a solution. $fg(x) = a\frac{x+1}{3} -1 = x\frac a3 + \frac{a-3}{3}$, and this is identically equal to $x$, so by comparing coefficients, $\frac a3 = 1$ and $\frac{a-3}{3} = 0$ which both have the consistent solution $a=3$. This is crucial.

If, on the other hand, your question has $f(x) = ax+1$, then $fg(x) = a\frac{x+1}{3}+1 = x\frac a3 + \frac{a+3}{3}$. If this is identically set equal to $x$, we get $\frac a3 = 1$ and $\frac{a+3}{3} = 0$ which have no consistent common solution. So there is no answer in this case.

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Since $$f(g(x))=x$$ for all $x$, it is true also for $x=0$: $${1\over 3}a-1=f({1\over 3})=f(g(0))=0$$

So $a=3$.

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  • $\begingroup$ should' it be f(g(x))=g(x)? $\endgroup$ – ten1o Sep 26 '18 at 9:41
  • $\begingroup$ No, it si $x$ ! $\endgroup$ – Aqua Sep 26 '18 at 9:42
  • $\begingroup$ why? thats not an identity function. $\endgroup$ – ten1o Sep 26 '18 at 9:42
  • $\begingroup$ Then which is it? $\endgroup$ – Aqua Sep 26 '18 at 9:43
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    $\begingroup$ Your solution before the edit gave a solution to a question that had no solution (due to the typo in the question). Your method of simply substituting a convenient numerical value failed to reveal the error in the question (which meant your original solution was also wrong). Right now (post-edit), your solution is correct, but it is not ideal because it remains susceptible to the same issue (not being able to pick up on when the question itself is wrong and no solution exists). $\endgroup$ – Deepak Sep 26 '18 at 12:45

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