Domain or range? I am doing some questions on contraction mapping and the first part of the question is: 

I am not sure what the question is asking for and how to do it.
Any help is hugely appreciated! thank you
 A: You want to find the range of $f(x)$ on the domain $[0,3]$; that is, the set $I=\{f(x)\,|\,x\in[0,3]\}$. Since the function is continuous on $[0,3]$, the set will be equal to
$$I=[\min_{x\in[0,3]} f(x),\max_{x\in[0,3]} f(x)].$$
To find the minimum and the maximum of the function, take a look at the sign of its derivative:
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2-2x}{2x+1}\right)=\frac{2(x^2+x-1)}{(2x+1)^2}$$
We see that the sign is determined by the sign of $x^2+x-1$, which is negative on $\left[0,\frac{\sqrt{5}-1}{2}\right]$, zero at $\frac{\sqrt{5}-1}{2}$, and positive on $\left[\frac{\sqrt{5}-1}{2},3\right]$.
Hence $$\min_{x\in[0,3]} f(x)=f\left(\frac{\sqrt{5}-1}{2}\right)=\frac{-3+\sqrt{5}}{2}$$ and $$\max_{x\in[0,3]} f(x)=\max\left\{f(0),f(3)\right\}=f(3)=\frac{3}{7},$$ so $I=\left[\frac{-3+\sqrt{5}}{2},\frac{3}{7}\right]$.
A: The question is asking you to find the set of all numbers of the form $f(x)$, where $x$ ranges over all numbers $0\le x\le 3$. A hint at how to do it: Start by computing a few such values (e.g., $f(0)$, $f(3)$, and perhaps a few more) to get a general feel for what is going on. Next try to sketch the graph of the function based on what you have found. That will at least get you started. Then, depending on how rigorous your arguments need to be, you'll need to refer to some theorems in order to establish your result. 
A: You are being asked to find the set of all numbers $f(x)$ for $0\le x\le 3$, where $f$ is the function 
$$f(x)=\frac{x^2-2x}{2x+1}\;.$$
For example, $f(0)=0$, $f(1)=-\frac13$, $f(2)=0$, and $f(3)=\frac37$, so $-\frac13$, $0$, and $\frac37$ are all in the set that you’re asked to find. Of course $[0,3]$ is an infinite set, so looking at specific values of $f(x)$ for $x\in[0,3]$ isn’t going to help much.
The function $f$ is continuous on its domain, which includes every real number except $-\frac12$ (why?), and $[0,3]$ is a closed interval in the domain of $f$, so you can use the extreme value and intermediate value theorems to answer the question. By the extreme value theorem you know that $f$ attains a minimum value on $[0,3]$ at some point $a\in[0,3]$ and a maximum value on $[0,3]$ at some point $b\in[0,3]$. By the intermediate value theorem you know that if $f(a)<y<f(b)$, there is some $x$ between $a$ and $b$ such that $f(x)=y$. Thus, the values of $f(x)$ for $x\in[0,3]$ are precisely the real numbers between $f(a)$ and $f(b)$, inclusive, i.e., the closed interval $[f(a),f(b)]$. That interval is what you’re asked to find.
