1. ## Piecewise defined function

I just read a thread in better ways to get help and thought I would make it easier by asking a specific question at a time as I get help. I don't really understand piecewise defined functions. For example:

Suppose that the function h is defined , for all real numbers, as follows.
h(x)= x if x is not equal to -3
5 if x =-3

I know I would have to plug in and solve for both, and make an x-y chart, but I don't get how to do that in this problem.. It still confuses me. I also need to know how to graph it. Any help would be very much appreciated.

Thank you,
Brittney

2. Originally Posted by Chinnie15
I just read a thread in better ways to get help and thought I would make it easier by asking a specific question at a time as I get help. I don't really understand piecewise defined functions. For example:

Suppose that the function h is defined , for all real numbers, as follows.
h(x)= x if x is not equal to -3
5 if x =-3

I know I would have to plug in and solve for both, and make an x-y chart, but I don't get how to do that in this problem.. It still confuses me. I also need to know how to graph it. Any help would be very much appreciated.

Thank you,
Brittney
What exactly are you trying to do with this piecewise defined function?

3. Graph h

4. Originally Posted by Chinnie15
Graph it
Just graph the function $\displaystyle h(x) = x$, and then at the point on the graph where $\displaystyle x = -3$, put an open dot (as the function is discontinuous at that point) and, also at $\displaystyle x = -3$ place a closed dot where $\displaystyle h(x) = 5$, since $\displaystyle h(x) = 5$ when $\displaystyle x = -3$.

5. Thanks! I am still lost though.. maybe it's just because it's late and I've been in this too long, but I still don't understand where I'm getting these values from. And how would I graph it? What would y be?

6. ## Piecewise functions

Hello Chinnie15
Originally Posted by Chinnie15
What would y be?
$\displaystyle y = h(x) =\left\{\begin{array}{c c}x, & \quad {x\ne -3}\\5, & \quad{x = -3}\\\end{array} \right.\$
I still don't understand where I'm getting these values from.
When you sketch the graph any function of $\displaystyle x$, it's up to you to decide what values of $\displaystyle x$ you want to include. The general answer is: any values that will show the general appearance of the graph. So in this case, the important thing is to get it right around $\displaystyle x = -3$. So choose values of $\displaystyle x$ from (say) $\displaystyle -10$ to (say) $\displaystyle +5$.
And how would I graph it?
In the way that Prove It has described.

7. Originally Posted by Chinnie15
Thanks! I am still lost though.. maybe it's just because it's late and I've been in this too long, but I still don't understand where I'm getting these values from. And how would I graph it? What would y be?
I'm sorry, but I don't understand where you are getting stuck...? You are given y = f(x), and a straight line (namely, y = x) to graph. You have exactly one other thing to do: Draw an open circle (indicating the missing point) on the straight line and a dot (indicating the odd point, (x, y) = (-3, 5)) off to the side of the line.

Please reply with clarification regarding what you have done and at what point you're grinding to a halt. Thank you!

8. Nevermind, I get it now. I have no idea how I forgot h(x) would be a strait line cutting diagonally right through the origin. Now that I actually remember that it's much easier. So that dot in the upper left would just stay there with no line connecting it?

9. Originally Posted by Chinnie15
Nevermind, I get it now. I have no idea how I forgot h(x) would be a strait line cutting diagonally right through the origin. Now that I actually remember that it's much easier. So that dot in the upper left would just stay there with no line connecting it?
Yes