# piecewise defined function

• Jul 1st 2010, 03:29 AM
reiward
piecewise defined function
Help me. Please check if I did the domain range and graph right. Thanks.
1.
http://img541.imageshack.us/img541/2280/math1.jpg

2.
http://img267.imageshack.us/img267/6237/math11.jpg
• Jul 1st 2010, 03:36 AM
Prove It
Quote:

Originally Posted by reiward
Help me. Please check if I did the domain range and graph right. Thanks.
1.
http://img541.imageshack.us/img541/2280/math1.jpg

2.
http://img267.imageshack.us/img267/6237/math11.jpg

1. Everything is correct except your graph.

2. The range is $\displaystyle \mathbf{R}\backslash \{6 \}$.
• Jul 1st 2010, 03:40 AM
reiward
Quote:

Originally Posted by Prove It
1. Everything is correct except your graph.

2. The range is $\displaystyle \mathbf{R}\backslash \{6 \}$.

1. Could you show me how to graph it. Thanks.

2. Oh, does that mean my graph is also wrong?
• Jul 1st 2010, 03:44 AM
Prove It
Quote:

Originally Posted by reiward
1. Could you show me how to graph it. Thanks.

2. Oh, does that mean my graph is also wrong?

1. The line should end at $\displaystyle x = 1$ with a QUADRATIC graph starting at $\displaystyle x = 1$.

2. The graph is almost right, you just need a hole at $\displaystyle (3, 6)$ on the line.
• Jul 1st 2010, 03:58 AM
reiward
Quote:

Originally Posted by Prove It
1. The line should end at $\displaystyle x = 1$ with a QUADRATIC graph starting at $\displaystyle x = 1$.

2. The graph is almost right, you just need a hole at $\displaystyle (3, 6)$ on the line.

1. Sorry to bother but can you draw it? I cant seem to picture it out. Im really not good at this sorry.

2. I get it. Ty! Oh. Is the range = [6] or not equal to 6?
• Jul 1st 2010, 06:12 AM
Prove It
Quote:

Originally Posted by reiward
1. Sorry to bother but can you draw it? I cant seem to picture it out. Im really not good at this sorry.

2. I get it. Ty! Oh. Is the range = [6] or not equal to 6?

$\displaystyle \mathbf{R}\backslash \{6\}$ means all the real numbers except 6.

For the graph, draw the graph of $\displaystyle y = x^2$. Then get rid of everything to the left of $\displaystyle x = 1$.