# domain and range

• Aug 26th 2008, 03:25 AM
sjenkins
domain and range
Find domain and range

f(x){-1,x<=1
3x+2,-1<x<1
7-2x, x>=1}

• Aug 26th 2008, 03:39 AM
Moo
Hello,
Quote:

Originally Posted by sjenkins
Find domain and range

f(x){-1,x<=-1
3x+2,-1<x<1
7-2x, x>=1}

For the domain, it's just the values x can take.

For the range... if x<=-1, then f(x)=-1. Thus the range contains -1.

f(x)=3x+2 if -1<x<1, that is to say -1<3x+2<5. This is part of the range too.

f(x)=7-2x if x>=1, that is to say 7-2x<=5. This actually gives the whole range.
• Aug 26th 2008, 03:49 AM
sjenkins
Functions have always been a particularly weak spot for me, I am not quite understanding. At first, I thought the domain was any number, but now I'm thinking -1,1. For the range, if 7-2x<=5, x would have to be 1 or less, right?
• Aug 26th 2008, 03:55 AM
Moo
Quote:

Originally Posted by sjenkins
Functions have always been a particularly weak spot for me, I am not quite understanding. At first, I thought the domain was any number, but now I'm thinking -1,1. For the range, if 7-2x<=5, x would have to be 1 or less, right?

$\displaystyle f(x)=\left\{ \begin{array}{lll} -1 \quad & \quad x \le -1 \\ 3x+2 \quad & -1<x<1 \\ 7-2x \quad & \quad x \ge 1 \end{array} \right.$

$\displaystyle x \le -1$ is like $\displaystyle ]-\infty~,-1]$, $\displaystyle -1<x<1$ is like $\displaystyle (-1~,1)$ and $\displaystyle x \ge 1$ is like $\displaystyle [1~,+\infty[$

So x can be any number in $\displaystyle \mathbb{R}$ (the set of real numbers).

Why would the range have been only -1 and 1 ? On the contrary, it's the values that would have been controversary because we're not sure the function is continuous at these points. In fact, it is.

You're told that if x>=1, then f(x)=7-2x. So find out what values 7-2x can take if x>=1. This gives 7-2x<=5
• Aug 26th 2008, 04:47 AM
sjenkins
Okay, maybe I should start from the beginning, for the given function, isn't the first number (or expression) the lowest number that can be used for the domain? And the second, the largest? That's why I picked -1 and 1 because in the first expression the number is -1 and in the last it is 1. I'm just really not sure. Isn't the range the result of what you put into the function?
• Aug 26th 2008, 04:54 AM
Moo
Quote:

Originally Posted by sjenkins
Okay, maybe I should start from the beginning, for the given function, isn't the first number (or expression) the lowest number that can be used for the domain? And the second, the largest? That's why I picked -1 and 1 because in the first expression the number is -1 and in the last it is 1. I'm just really not sure. Isn't the range the result of what you put into the function?

Yes. This means that you want the different values that f(x) can have.

For any value of x that is inferior to -1, the function is -1
For any value of x that is superior to -1 and inferior to 1, we have :
$\displaystyle -1<x<1 \implies -3<3x<3 \implies -1<\underbrace{3x+2}_{f(x) \text{ for } -1<x<1}<5$. So the different values of f(x) while -1<x<1 are all the values within -1 and 5.

So far, the range is $\displaystyle [-1~,5)$.

For any value of x that is superior to 1, we have :
$\displaystyle x \ge 1 \implies 2x \ge 2 \implies -2x \le -2 \implies \underbrace{7-2x}_{f(x) \text{ for } x \ge 1} \le 5$. So the different values of f(x) while $\displaystyle x \ge 1$ are all the values inferior to 5.

So we have $\displaystyle f(x) \le 5$ if $\displaystyle x \ge 1$, which $\displaystyle f(x) \in (-\infty~,5]$

The range will be the union of these two intervals :
$\displaystyle R=[-1~,5) \cup (-\infty~,5]=(-\infty~,5]$

Is it clear enough ? (Worried)
• Aug 26th 2008, 05:17 AM
sjenkins
Where did the 5 come from in -1<3x+2<5? I'm sorry, I really am trying to understand this.
• Aug 26th 2008, 05:27 AM
Moo
Quote:

Originally Posted by sjenkins
Where did the 5 come from in -1<3x+2<5?

I added 2 to each side (3+2), like I did for 7-2x :)

Quote:

I'm sorry, I really am trying to understand this.
You don't have to be sorry, it's a good thing ! (Wink)