Posted by Lori on Friday, March 28, 2008 at 5:01pm.
Only the middle one counts as I understand your question.
x is not less than one. It is three
When x is 3, then 3<= x
x is not more than eight,. It is three.
so for x = 3 , f(x) = 3 x = 9
But the reason I am having trouble is because the middle one does not work.
The first part works, but the second part does not.
3<=x<=8=3(3)=9
8=3(3)=9 (?)
That just means that between x = 3 and x = 8 (including 3 and 8)
You can use
y = 3 x
if x were 4, you could say y = 3*4 = 12
BUT
if x is 9 , you are not allowed to use this relation and have to use some other one.
In this particular case, you would use
y = 3 - 5x
if x were 9, which is greater than 8
Thank-you for explaining it.
Related Questions
Algebra HELP! - I cannot find an example and am confused about what to do with ...
algebra - How do you solve this math problem? h= -8t^2+40t then it says: when t=...
Algebra HELP! - I cannot find an example and am confused about what to do with ...
Math SOMEONE PLEASE PLEASE PLEASE PLEASE HELP - Find sin(θ), cos(&#...
Algebra - I am having trouble with this problem. I have done it several times ...
General - Please understand that I am not upset with anyone or whatever I just ...
Algebra II repost- please check - I posted these a little while ago but this ...
Calculus - The problem is lim --> 2 for g(x) which is 2((x-2/(squareroot ...
calculus - The problem is lim --> 2 for g(x) which is 2((x-2/(squareroot ...
calculus - The problem is lim --> 2 for g(x) which is 2((x-2/(squareroot ...
For Further Reading