1. ## Limit

Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets.

would someone give me a brief explanation on how to solve such equations?
(note: these are greatest integer brackets)

2. Originally Posted by Calc1
Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets.

would someone give me a brief explanation on how to solve such equations?
(note: these are greatest integer brackets)

$\displaystyle g(x)=\left\{ \begin{array}{cc} 0 & \text{if } -3 \leq x \leq 3 \end{array} \right.$

I can't get the latex right but it looks to me like the function is zero for all x

1-x is the additive invers of x-1

3. Originally Posted by VonNemo19
$\displaystyle g(x)=\left\{\begin{array}{cc}0,&\mbox{if}{-3\leq{x}\leq{3}}\end{array}\right$

I can't get the latex right but it looks to me like the function is zero for all x

1-x is the additive invers of x-1
Counterexample: g(2.5)=[1.5]+[1-2.5]
=1+[-1.5]
=1+(-2)
=-1

4. i understand the greatest integer theory but i still don't get how it would be solved in this question

as you stated g(2.5) = -1

ok i know how you got that but how is it exactly related to the question?

5. Originally Posted by Calc1
Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets.

would someone give me a brief explanation on how to solve such equations?
(note: these are greatest integer brackets)
let

f(x)=[x-1]
h(x)=[1-x] $\displaystyle x\in [-3,3]$

$\displaystyle f(x)=\begin{Bmatrix} -4&...if...-3\leq x<-2\\ -3&...if...-2\leq x<-1\\ -2&...if...-1\leq x<0\\ -1&...if...0\leq x<1\\ 0&...if...1\leq x<2\\ 1&...if...2\leq x<3\\ 2&...if...x=3 \end{Bmatrix}$

$\displaystyle h(x)=\begin{Bmatrix} 4&...if...x=-3\\ 3&...if...-3< x\leq -2\\ 2&...if...-2< x\leq -1\\ 1&...if...-1< x\leq 0\\ 0&...if...0< x\leq 1\\ -1&...if...1< x\leq 2\\ -2&...if...2< x\leq 3 \\ \end{Bmatrix}$

$\displaystyle g(x)=\begin{Bmatrix} 0&...if...x=-3\\ -1&...if...-3< x< -2\\ 0&...if...x=-2\\ -1&...if...-2< x< -1\\ 0&...if...x=-1\\ -1&...if...-1< x< 0\\ 0&...if...x=0\\ \end{Bmatrix}$
$\displaystyle ............\begin{Bmatrix} -1&...if...0< x< 1\\ 0&...if...x=1\\ -1&...if...1< x< 2\\ 0&...if...x=2\\ -1&...if...2< x<3\\ 0&...if...x=3\\ \end{Bmatrix}$ since we can write f(x) and h(x) like below then find the sum of f(x) and h(x) because g(x)=f(x)+h(x)

$\displaystyle f(x)=\begin{Bmatrix} -4&...if... x=-3\\ -4&...if...-3< x<-2\\ -3&...if...x=-2\\ -3&...if...-2< x<-1\\ -2&...if...x=-1\\ -2&...if...-1< x<0\\ -1&...if...x=0\\ \end{Bmatrix}$
$\displaystyle ...........\begin{Bmatrix} -1&...if...0< x<1\\ 0&...if...x=1\\ 0&...if...1< x<2\\ 1&...if...x=2\\ 1&...if...2< x<3\\ 2&...if...x=3 \end{Bmatrix}$

$\displaystyle h(x)=\begin{Bmatrix} 4&...if... x=-3\\ 3&...if...-3< x<-2\\ 3&...if...x=-2\\ 2&...if...-2< x<-1\\ 2&...if...x=-1\\ 1&...if...-1< x<0\\ 1&...if...x=0\\ \end{Bmatrix}$
$\displaystyle ...........\begin{Bmatrix} 0&...if...0< x<1\\ 0&...if...x=1\\ -1&...if...1< x<2\\ -1&...if...x=2\\ -2&...if...2< x<3\\ -2&...if...x=3 \end{Bmatrix}$

it is clear or not

6. you can take 1 from the greatest integer function since the factor of x is one

so you can write g(x) like this

$\displaystyle g(x)=[x-1]+[1-x]=[x]-1+1+[-x]=[x]+[-x]$

but the result will be the same as I solve before

7. Originally Posted by alexmahone
Counterexample: g(2.5)=[1.5]+[1-2.5]
=1+[-1.5]
=1+(-2)
=-1
I've made three mistakes within the last 24 hours. I must be losing it.

8. Thanks Amer, it makes sense to me now