Given $f(x)=-x^2+1$ and $g(x)=\sqrt{x+1}$, find $k(x)=(g\circ f)(x)$? Given $f(x)=-x^2+1$ and $g(x)=\sqrt{x+1}$, find $k(x)=(g \circ f)(x)$?
Following the step's my teach told me to do this type of equation I did this... I feel like I'am not showing enough work and I'am going to college next year and I want to get full mark I'am not even sure If i'am correct. please help me learn how to do this...
My solution : 
$$(g \circ f)(x) = 
g(f(x)) = 
\sqrt{(-x^2 + 1) + 1} = 
\sqrt{2 - x^2}$$
My apologizes for not formatting this correctly.. 
 A: Your solution is correct. $\textbf{}$
A: $\text{Just compare, few moments back, I found a function}\\$
$$\text{$f(x) = \sqrt x$}\\$$
$$\text{g(x)} =x^2\\$$
$\text{domain of $f(x)=\sqrt x$ is all non-negative real numbers, while}\\$
$\text{domain of $g(x) = x^2$ is all the real numbers,}\\$
$\text{the composed function is $(g \circ f)(x) = g(f(x))=(\sqrt x)^2=x$}\\$
$\text{Now, x would nomally have the domain of all real numbers,}\\$
$\text{But, because it is a composed function, we must also consider $f(x)$,}\\$ 
$\text{So, the domain is all non-negative real numbers}\\$
$\text{$(def:-$ where $Domain$ of a function is the set of input argument value,}\\$
$\text{$\qquad$ for which the function is defined.)}\\$
A: As several people in the comments pointed out, that is completely correct.
As far as showing work goes, what you've shown is great.  A really important part is the parenthesis that you have in your $3$rd part of the equality--this shows that you understand that the entirety of $f(x)$ replaces the $x$ in $g(x)$.
