I seem to be struggling with part B of this assignment.
I plot the function into every value of $x$, yet it does not help me to solve it at all! Is there a different way to do this? Is the substitution of $x$ by $f(x)$ necessary, or does a less time consuming way exist(perhaps doing something to the previously found domain and range of $f$)?
Appreciate the help.
Hint (a): to find the range of $f$, let $y = f(x) = 2 - \frac{x+1}{x^2+2x+2}$. Then:
$$y(x^2+2x+2)=2(x^2+2x+2) - (x+1)$$
$$(y - 2)x^2 + (2 y - 3) x + 2y -3 = 0$$
The range of $f$ is the range of values $y$ for which the above has real roots.
$$\Delta = (2y-3)^2-4(y-2)(2y-3) = -(2y-3)(2y-5) \;\;\ge\;0$$
Hint (b): once you determined at step (a) that the range of $f$ is $f(\mathbb{R})=[\frac{3}{2},\frac{5}{2}]$, the range of $f \circ f$ will be $f(f(\mathbb{R}))=f([\frac{3}{2},\frac{5}{2}])$. It helps to notice that $f$ is increasing for $x \ge 0$, which becomes more apparent if you write it as:
$$f(x) = 2 - \frac{1}{(x+1)+\cfrac{1}{(x+1)}}$$