MHB Solving Piecewise Functions with U(x) - U(x-1): Explanation and Graph

[paulmdrdo1]

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]

Re: piecewise function questions

Hello and welcome to MHB, paulmdrdo! (Wave)

We are given:

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

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

 

[paulmdrdo1]

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]

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.

 

[paulmdrdo1]

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]

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 $[0,1]$?

 

[paulmdrdo1]

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]

What did you find when you did the subtraction?

 

[paulmdrdo1]

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]

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?

 

[paulmdrdo1]

This is what I did:

https://www.physicsforums.com/attachments/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]

First, do you see that on:

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

$$(0,1)$$ we have: $$g(x)=1$$

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

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:

$$U(0)-U(0-1)=1-0=1$$

$$U(1)-U(1-1)=1-1=0$$

Now, this means we may amend our intervals as:

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

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

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

and so we may write:

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

 

[paulmdrdo1]

First, do you see that on:

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

$$(0,1)$$ we have: $$g(x)=1$$

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

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:

$$U(0)-U(0-1)=1-0=1$$

$$U(1)-U(1-1)=1-1=0$$

Now, this means we may amend our intervals as:

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

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

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

and so we may write:

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

thank you very much mark! you're very generous in answering my questions! i now fully understand it! thanks!