If: $$g\Big(\frac{x-1}{x-1}\Big)=2x, f(x)=\frac {\sqrt x-1}{\sqrt x +1}, h(x)=x^2, p(x)=f(h(x))$$

Then $g(p(3))=?$

I don't know if i can simplify the first function to be $g(1)=2x$, but i don't think so because $x \ne 1$. In this step is where im stuck.

I only need to know $g(\frac{1}{2})$ and i'm done, but i don't know what to do.

Any hints?

  • 1
    $\begingroup$ Where is this puzzle from? $\endgroup$ – Mason Jul 14 '18 at 4:05
  • $\begingroup$ There is some $x$ such that $g(1)=2x$? I mean taking $x=5$ and $10$ this says that $2\times 5 =g(1)=2 \times 10$ but that means that $5=10$ and that would be... bad. $\endgroup$ – Mason Jul 14 '18 at 4:08
  • $\begingroup$ I know, i said that i don't think its possible. And the exercise isn't a puzzle, is in a "advanced" guide from a "college prep course" in my country. This is one of the exercise that i can't do. $\endgroup$ – Rodrigo Pizarro Jul 14 '18 at 4:28
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    $\begingroup$ I think it is a typo. It is a good precalc prep problem if we say that $g(\frac{x+1}{x-1})=2x$. Maybe a + has faded into a minus? $\endgroup$ – Mason Jul 14 '18 at 4:45

$f(x)=\frac {\sqrt x-1}{\sqrt x +1}$

$ h(x)=x^2$

$ p(x)=f(h(x))$

Now let us unpack $p(x)=\frac{x-1}{x+1}$

Then $p(3)=-\frac{2}{4}=-\frac{1}{2}$

Then you must compute $g(-\frac{1}{2})$ but my best interpretation is that there is some value for $x$ such that $g(1)=2x$. And that's not very helpful in interpreting $g(-1/2)$. Let us say instead that for all values of $x$ we have that $g(\frac{x+1}{x-1})=2x$.

What can we do from here?

  • $\begingroup$ I think it its a typo, i will ask my teacher soon. Thanks. $\endgroup$ – Rodrigo Pizarro Jul 14 '18 at 5:21

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