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)

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

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] $x\in [-3,3]$

$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}

$

$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}$

$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}$

$
............\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)

$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}$

$
...........\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}

$

$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}$

$
...........\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

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

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