For the function
_____{x+3 if -5≤x<2
f(x)={x^2 if 2≤x≤4
I have to sketch the graph, determine the domain and range of f(x), and determine f(2). Help please?
$\displaystyle f(x)=\left\{\begin{array}{ll}x+3 & ,-5\leq x<2\\x^2 & ,2\leq x\leq 4\end{array}\right.$
The domain is $\displaystyle [-5,4]$.
To find the range we can use the following statement abou function:
If $\displaystyle f:A\to B$ and $\displaystyle X, \ Y$ are two subsets of A, then
$\displaystyle f(X\cup Y)=f(X)\cup f(Y)$
In this case let $\displaystyle X=[-5,2), \ Y=[2,4]$
We have $\displaystyle f(X)=[-2,5), \ f(Y)=[4,16]$.
Then the range is $\displaystyle [-2,5)\cup[4,16]=[-2,16]$
Then let's do it in another way.
$\displaystyle -5\leq x<2$
Add 3 to all members: $\displaystyle -2\leq x+3<5$
Then, if $\displaystyle x\in[-5,2)$ then $\displaystyle f(x)\in[-2,5)$
$\displaystyle 2\leq x\leq 4$
Square all members: $\displaystyle 4\leq x^2\leq 16$
Then, if $\displaystyle x\in[2,4]$ then $\displaystyle f(x)\in[4,16]$.
So, the range is $\displaystyle [-2,5)\cup[4,16]=[-2,16]$
Is it better now?
Sketch y = x + 3, and then erase everything before x = -5 and after x = 2. Make sure to draw a filled-in circle for the left-hand endpoint and an "open" circle for the right-hand endpoint.
Then sketch y = x^2, and erase everything before x = 2 and after x = 4. Make sure to draw filled-in circles for each of the endpoints.
The domain is given: it's the x-values for which the function is defined.
To find the range, look at your graph. Which y-values are covered by this graph? (If you "collapsed the graph sideways onto the y-axis, which portions would be covered?)
The function is defined for x = 2. So look at the function rule, find the half which is defined for x = 2, and plug 2 in for x in that half's rule.