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$$f(x)= \begin{cases} 2x & ; x < 0 \\ \sqrt{x} & ; 0 \leq x \leq 1 \\ (x-1)^2+1 & ; x > 1 \end{cases} $$ $$g(x)= \begin{cases} x^2 & ; x \leq 1 \\ 1 &; x > 1 \end{cases} $$ Hello, firstly I would like to thank everyone who is reading this post. I need to say, that I already read all of the similar titles and tried to understand why the solution is such and what steps you need to take to get to that solution. However, I can't solve this problem on my own. all I am trying to do is composition of $f(g(x))$.

What I think I should be doing is: firstly look at what happens with $x \leq 1$ in $g(x)$. Then I input $g(x)$ which in case of $x\leq1$ is $x^2$ into $f(x)$. After I do that I am pretty much lost on what to do, let alone the fact I don't understand what ranges to write down for particular $f(g(x))$.

I am not an native speaker, as you can probably tell, but I hope the one reading this question can understand it enough to help me.

Thank you in advance,

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  • $\begingroup$ Try to find the range of $g(x)$ on each interval where it's defined. What does that imply on the domain of $f(g(x))$? $\endgroup$ – big-lion Jun 27 '18 at 13:36
  • $\begingroup$ Similar question, might be helpful. math.stackexchange.com/q/317416/502324 $\endgroup$ – big-lion Jun 27 '18 at 21:47
  • $\begingroup$ @big-lion Hey, I still need help. $\endgroup$ – tone tinkovič Jul 2 '18 at 11:47
  • $\begingroup$ Sure. What have you tried so far? What did you achieve? Did you understand what me and Bill hunter? $\endgroup$ – big-lion Jul 2 '18 at 11:48
  • $\begingroup$ @big-lion I posted what I did so far bellow? if you can't see I am actually trying to solve it for gof atm. $\endgroup$ – tone tinkovič Jul 2 '18 at 12:17
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As big-lion hinted:

  • Start by finding the values of $x$ where $g(x)<0$; those where $0\leq g(x)\leq 1$ and those where $g(x)>1$.

  • Then, replace $y$ by the value of $g(x)$ in the expression of $f(y)$.

Give it a try and show us what you come up with if you need help.

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  • $\begingroup$ Something like this if I understood correctly?$$f(x)= \begin{cases} 2(g(x)) & ; g(x) < 0 \\ \sqrt{g(x)} & ; 0 \leq g(x) \leq 1 \\ (g(x)-1)^2+1 & ; g(x) > 1 \end{cases} $\endgroup$ – tone tinkovič Jun 27 '18 at 18:29
  • $\begingroup$ I need help. I am lost on what to do or how to proceed. @bill-oharan $\endgroup$ – tone tinkovič Jun 28 '18 at 12:18
  • $\begingroup$ That's the spirit but then, what is $g(x)<0$ in terms of $x$? And what is $2g(x)$ then? $\endgroup$ – Bill O'Haran Jun 28 '18 at 12:24
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    $\begingroup$ ohh I think I see where I misspoke. I thought we were inputing g(x) into ranges of f(x). $\endgroup$ – tone tinkovič Jun 28 '18 at 14:33
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    $\begingroup$ @bill-oharan yes of course I will construct something appealing for the one just looking for answer :) $\endgroup$ – tone tinkovič Jun 28 '18 at 16:51

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