$$f(x)= \begin{cases} 2x & ; x < 0 \\ \sqrt{x} & ; 0 \leq x \leq 1 \\ (x-1)^2+1 & ; x > 1 \end{cases} $$ $$g(x)= \begin{cases} x^2 & ; x \leq 1 \\ 1 &; x > 1 \end{cases} $$ 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,
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.