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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

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!

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

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!

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- #2

Hello and welcome to MHB, paulmdrdo!

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.

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- May 13, 2013

- 386

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

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!

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- #4

- Thread starter
- #5

- May 13, 2013

- 386

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?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.

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- May 13, 2013

- 386

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

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- May 13, 2013

- 386

what i did wasmark 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!

(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.

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- #11

- May 13, 2013

- 386

based on the third graph the new function is defined by,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?

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.

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- #12

\(\displaystyle (-\infty,0)\) 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:

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

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- #13

- May 13, 2013

- 386

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

\(\displaystyle (-\infty,0)\) 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:

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