find this function value  Which interval is $\,x\,$ in?  so, use this rule...  solution 

$f(1)$  $1\in [1,1)$  $x+3\,$ (first row)  $f(1) = (1) + 3 = 4$ 
$f(0)$  $0\in [1,1)$  $x+3\,$ (first row)  $f(0) = (0) + 3 = 3$ 
$f(0.9)$  $0.9\in[1,1)$  $x+3\,$ (first row)  $f(0.9) = (0.9) + 3 = 2.1$ 
$f(1)$  $1\in [1,2)\,$  $5\,$ (second row)  $f(1) = 5$ 
$f(1.99999)$  $1.99999 \in [1,2)$  $5\,$ (second row)  $f(1.99999) = 5$ 
$f(2)$  $2 \in [2,\infty)$  $x^2\,$ (third row)  $f(2) = 2^2 = 4$ 
$f(\pi)$  $\pi \in [2,\infty)$  $x^2\,$ (third row)  $f(\pi) = \pi^2$ 
If forced to read the function
$\displaystyle
\cssId{s51}{
f(x) =
\cases {
x + 3 &\text{if }\quad 1 \le x < 1\cr
5 &\text{if }\quad 1 \le x < 2\cr
x^2 &\text{if }\quad x\ge 2
}}
$
aloud, it is typically read from lefttoright and toptobottom.
One possible reading could go like this:
Here are additional things you should be able to do with a piecewisedefined function:
The domain is easy to get from the formula alone.
Just ‘put together’ (union) the input sets from all the rows.
Combined, these are the values of $\,x\,$ that $\,f\,$ knows how to act on.
top row  $[1,1)$  
second row  $\,[1,2)\,$  
third row  $\,[2,\infty)\,$  
union, to get the domain of $\,f\,$  $\text{dom}(f) = [1,\infty)\,$ 
By the way, the range of $\,f\,$ is nowhere near as easy to get from the formula.
Once you have a graph, though, the range is easy to obtain:
just use the ‘collapse the graph into the $\,y\,$axis’ method,
as described in this earlier lesson.
In this example, three rules are needed to fully understand what happens at and near $\,x\,$:
to be a function, each allowable input must have exactly one output. 
Next, we graph:
$$
\cssId{s105}{f(x) =
\cases {
x + 3 &\text{if }\quad 1 \le x < 1\cr
5 &\text{if }\quad 1 \le x < 2\cr
x^2 &\text{if }\quad x\ge 2
}}
$$
The ‘pieces’ may be graphed in any order.
Here, the choice is made to move from lefttoright on the number line (from toptobottom in the formula).
On $\color{red}{[1,1)}$, $\,f\,$ is linear,
since $\,\color{red}{x+3}\,$ is of the form $\,mx + b\,$. The endpoint $\,1\,$ is included in the interval, indicated by the use of the bracket ‘$\,[\,$’ in interval notation. The endpoint $\,1\,$ is excluded from the interval, indicated by the use of the parenthesis ‘$\,)\,$’ in interval notation.
Important: the point $\,(1,2)\,$ is not a point on the graph of $\,f\,$. This hollow dot is used to show limiting behavior: as the inputs approach $\,1\,$ from the lefthand side, the outputs approach $\,2\,$. 

On $\color{blue}{[1,2)}$, $\,f\,$ is constant with value $\color{blue}{5}$, so it graphs as a horizontal segment.
Important: the point $\,(2,5)\,$ is not a point on the graph of $\,f\,$. This hollow dot is used to show limiting behavior: as the inputs approach $\,2\,$ from the lefthand side, the outputs approach $\,5\,$. 

On $\color{green}{[2,\infty)}$, $\,f\,$ is quadratic,
since $\,\color{green}{x^2}\,$ is of the form $\,ax^2 + bx + c\,$. Thus, it graphs as a piece of a parabola on this interval. Think about what the entire graph of $\,y = x^2\,$ looks like (it is lightly dashed here), and draw in the portion for the inputs $[2,\infty)\,$.


Here is the graph of the piecewisedefined function $\,f\,$, colorcoded with its graph: $$ \cssId{s159}{f(x) = \cases { \color{red}{x + 3} &\text{if }\quad 1 \le x < 1\cr \color{blue}{5} &\text{if }\quad 1 \le x < 2\cr \color{green}{x^2} &\text{if }\quad x\ge 2 }} $$ 
It remains only to get the formula for the linear section:
$$
\cssId{s187}{
f(x) =
\cases {
x + 3\color{red}{\,,} &\text{if }\quad 1 \le x < 1\cr
5\color{red}{\,,} &\text{if }\quad 1 \le x < 2\cr
x^2\color{red}{\,,} &\text{if }\quad x\ge 2
}}
$$ 
$$
\cssId{s189}{f(x) =
\cases {
x + 3\,, &\text{if }\quad 1 \le x < 1\color{green}{\,,}\cr
5\,, &\text{if }\quad 1 \le x < 2\color{green}{\,,}\cr
x^2\,, &\text{if }\quad x\ge 2
}}
$$

$$
\cssId{s191}{f(x) =
\cases {
x + 3\,, &\text{for}\quad 1 \le x < 1\cr
5\,, &\text{for}\quad 1 \le x < 2\cr
x^2\,, &\text{for}\quad x\ge 2
}}
$$

With beautiful typesetting (including all the words ‘ if ’ lined up vertically), this author prefers the simplest formatting, with no extra punctuation.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
