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[SOLVED] Piecewise function questions

paulmdrdo

Active member
May 13, 2013
386
given function U(x)

U(x) = { 0, if x<0
{1, if x>=0

define the fuction piecewise and graph.
* U(x) - U(x-1)

how to solve this? please explain! thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: piecewise function questions

Hello and welcome to MHB, paulmdrdo! (Wave)

We are given:

\(\displaystyle U(x)=\begin{cases}1 & x<0\\ 0 & x\ge0 \\ \end{cases}\)

I would graph this, then graph $U(x-1)$ beneath it, recogninzing that this is just the original graph shifted 1 unit to the right. Now, subtract the lower graph from the upper graph, to create a third graph beneath the other two, representing $g(x)=U(x)-U(x-1)$. From this, you should be able to define $g(x)$ as a piecewise defined function.

Show us what you find. :D
 

paulmdrdo

Active member
May 13, 2013
386
Re: piecewise function questions

given function U(x)

U(x) = { 0, if x<0
{1, if x>=0

define the fuction piecewise and graph.
* U(x) - U(x-1)

how to solve this? please explain! thanks!
thank you mark! now i have an idea how to solve this graphically. but is there a way to solve this algebraically?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: piecewise function questions

I think doing this graphically is the easiest method. Otherwise, you will wind up doing essentially the same thing, but without the benefit of a graph to make it more clear.
 

paulmdrdo

Active member
May 13, 2013
386
Re: piecewise function questions

I think doing this graphically is the easiest method. Otherwise, you will wind up doing essentially the same thing, but without the benefit of a graph to make it more clear.
oh no! please bear with my stupidity. i don't know how to subtract the graph visually. what am I going to subtract, the x-values or the y-values?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: piecewise function questions

You will subtract the function values (the $y$-values) for all real $x$. Can you see what the difference $g(x)$ is outside of $[0,1]$?
 

paulmdrdo

Active member
May 13, 2013
386
Re: piecewise function questions

You will subtract the function values (the $y$-values) for all real $x$. Can you see what the difference $g(x)$ is outside of $[0,1]$?
mark i tried subtracting the coordinates of the lower graph from the coordinates of upper graph. but i don't know if that's correct. can you give me some examples. thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What did you find when you did the subtraction?
 

paulmdrdo

Active member
May 13, 2013
386
Re: piecewise function questions

mark i tried subtracting the coordinates of the lower graph from the coordinates of upper graph. but i don't know if that's correct. can you give me some examples. thanks!
what i did was

(0,1) - (0,0) = (0, 1) - included
(1,1) - (1,1) = (0,0) - not included
and so on...
still i'm not sure about this. please can you show me what you did to solve this.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This is what I did:

paulmdrdo.jpg

Now, after you are make sure you see what I did, can you use the bottom graph, and write the piecewise function definition for it?
 

paulmdrdo

Active member
May 13, 2013
386
This is what I did:

View attachment 856

Now, after you are make sure you see what I did, can you use the bottom graph, and write the piecewise function definition for it?
based on the third graph the new function is defined by,

U(x)-U(x-1) = {0, x>=1; 1, 0<=x<1; 0, x<0

but i'm having a hard time subtracting the graph visually.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
First, do you see that on:

\(\displaystyle (-\infty,0)\) we have: \(\displaystyle g(x)=0\)

\(\displaystyle (0,1)\) we have: \(\displaystyle g(x)=1\)

\(\displaystyle (1,\infty)\) we have \(\displaystyle g(x)=0\)

So, I would draw these with open end-points at $x=0,\,1$ Then, to see which to fill in and which to leave open, observe that:

\(\displaystyle U(0)-U(0-1)=1-0=1\)

\(\displaystyle U(1)-U(1-1)=1-1=0\)

Now, this means we may amend our intervals as:

\(\displaystyle (-\infty,0)\) we have: \(\displaystyle g(x)=0\)

\(\displaystyle [0,1)\) we have: \(\displaystyle g(x)=1\)

\(\displaystyle [1,\infty)\) we have \(\displaystyle g(x)=0\)

and so we may write:

\(\displaystyle g(x)=\begin{cases}0 & x<0 \\ 1 & 0\le x<1 \\ 0 & 1\le x \\ \end{cases}\)
 

paulmdrdo

Active member
May 13, 2013
386
First, do you see that on:

\(\displaystyle (-\infty,0)\) we have: \(\displaystyle g(x)=0\)

\(\displaystyle (0,1)\) we have: \(\displaystyle g(x)=1\)

\(\displaystyle (1,\infty)\) we have \(\displaystyle g(x)=0\)

So, I would draw these with open end-points at $x=0,\,1$ Then, to see which to fill in and which to leave open, observe that:

\(\displaystyle U(0)-U(0-1)=1-0=1\)

\(\displaystyle U(1)-U(1-1)=1-1=0\)

Now, this means we may amend our intervals as:

\(\displaystyle (-\infty,0)\) we have: \(\displaystyle g(x)=0\)

\(\displaystyle [0,1)\) we have: \(\displaystyle g(x)=1\)

\(\displaystyle [1,\infty)\) we have \(\displaystyle g(x)=0\)

and so we may write:

\(\displaystyle g(x)=\begin{cases}0 & x<0 \\ 1 & 0\le x<1 \\ 0 & 1\le x \\ \end{cases}\)
thank you very much mark! you're very generous in answering my questions! i now fully understand it! thanks!