Piecewise Functions and Domains

In summary, the conversation discussed two questions and their solutions. For the first question, the solution involved a piecewise function with different values depending on the interval. For the second question, the solution also involved a piecewise function with different royalties rates depending on the number of copies sold. The conversation also included some clarifications and corrections on the solutions.
  • #1
ardentmed
158
0
Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:
08b1167bae0c33982682_2.jpg


For the first one, I got a dotted graph (with 1,4 as the initial value) and then precipitous jumps in the graph. So I said that:
f(x)=1 if x is over (0,1]
f(x)=5 if x is over (1,1.2]
..
f(x)=9 if x is over (1.8,2.0]

As for the second one, I stated R(x) as a piecewise function, where:

R(x) = 15(0.10x) when x <= 10,000
R(x) = 15(0.15x) when x is over (10,000, 20,000]
R(x) = 15(0.20x) when x > 20,000

Which simplifies to:
R(x) = 1.5x when x <= 10,000
R(x) = 2.25x + 15,000 when x is over (10,000, 20,000]
R(x) = 3x + 60,000 when x > 20,000Am I on the right track?

Thanks in advance.
 
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  • #2
For question 1), instead of defining the interval (1, 2] into 5 different portions, I would define them as one. I.e if \(\displaystyle 1 < x \le 2\). f(x) = 4 + 5(x-1)
4 is for the first kilometer, and (x-1) calculates the distance traveled past the first kilometer multiplied by a number (in this case, 5) to get our desire value.

Question 2) is a different type of question than one.

You are correct for the first portion. If each paperback sells for $15, then we must multiply the profit (15x) to 10%.

R(x) = 0.10(15x) when [0, 10,000]


However, for (10,000, 20,000], only the number of copies sold OVER 10,000 is 15% royalties, the first 10,000 remains at 10% royalties!

So for (10,000, 20,000], we need to add the revenue from the first portion, [0, 10,000], to 15% royalties of the copies over 10,000.

R(x) = 0.10(15)(10,000) + (0.15)(15)(x-10000) when (10,000, 20,000]


Can you now do the last region when x > 20,000?
 
Last edited:
  • #3
ardentmed said:
For the first one, I got a dotted graph (with 1,4 as the initial value) and then precipitous jumps in the graph. So I said that:
f(x)=1 if x is over (0,1]
Did you mean $f(x)=4$?

Rido12 said:
For question 1), instead of defining the interval (1, 2] into 5 different portions, I would define them as one. I.e if \(\displaystyle 1 < x \le 2\). f(x) = 4 + 5(x-1)
This is not correct since the actual $f(x)$ is not continuous on $(1,2]$. The OP's answer is correct. The formula can also be written as follows
\[
f(x)=
\begin{cases}
4&0<x\le1\\
4+\lceil 5(x-1)\rceil&1<x\le 2
\end{cases}
\]
where $\lceil\cdot\rceil$ is the ceiling (round up) function.

[GRAPH]wat9ga1hpc[/GRAPH]
 
  • #4
Evgeny.Makarov said:
Did you mean $f(x)=4$?

This is not correct since the actual $f(x)$ is not continuous on $(1,2]$. The OP's answer is correct. The formula can also be written as follows
\[
f(x)=
\begin{cases}
4&0<x\le1\\
4+\lceil 5(x-1)\rceil&1<x\le 2
\end{cases}
\]
where $\lceil\cdot\rceil$ is the ceiling (round up) function.

You are absolutely correct. I forgot the ceiling function.
 
  • #5
Evgeny.Makarov said:
Did you mean $f(x)=4$?

This is not correct since the actual $f(x)$ is not continuous on $(1,2]$. The OP's answer is correct. The formula can also be written as follows
\[
f(x)=
\begin{cases}
4&0<x\le1\\
4+\lceil 5(x-1)\rceil&1<x\le 2
\end{cases}
\]
where $\lceil\cdot\rceil$ is the ceiling (round up) function.
Yes, I meant to say 4. So is my answer correct?

Thanks in advance for the help guys. I really appreciate it.
 
  • #6
Rido12 said:
For question 1), instead of defining the interval (1, 2] into 5 different portions, I would define them as one. I.e if \(\displaystyle 1 < x \le 2\). f(x) = 4 + 5(x-1)
4 is for the first kilometer, and (x-1) calculates the distance traveled past the first kilometer multiplied by a number (in this case, 5) to get our desire value.

Question 2) is a different type of question than one.

You are correct for the first portion. If each paperback sells for $15, then we must multiply the profit (15x) to 10%.

R(x) = 0.10(15x) when [0, 10,000]


However, for (10,000, 20,000], only the number of copies sold OVER 10,000 is 15% royalties, the first 10,000 remains at 10% royalties!

So for (10,000, 20,000], we need to add the revenue from the first portion, [0, 10,000], to 15% royalties of the copies over 10,000.

R(x) = 0.10(15)(10,000) + (0.15)(15)(x-10000) when (10,000, 20,000]


Can you now do the last region when x > 20,000?
For the second question, for the last part (x > 20,000), I computed:

R(x) = 0.10(15)(10,000) + (0.15)(15)(10,000) + (3x + 60,000)(x-20,000) when (20,000, infinity][/B]

Am I close? Thanks again for the help.
 
  • #7
ardentmed said:
For the second question, for the last part (x > 20,000), I computed:

R(x) = 0.10(15)(10,000) + (0.15)(15)(10,000) + (3x + 60,000)(x-20,000) when (20,000, infinity][/B]

Am I close? Thanks again for the help.

Can you clarify on how you got \(\displaystyle (3x + 60,000)\)?
 
  • #8
Rido12 said:
Can you clarify on how you got \(\displaystyle (3x + 60,000)\)?

Having gone through my work, I can't even recall why I used 3x + 60,000, it must have been a simplification of one of the functions I came up with earlier, perhaps after accounting for the 20,000 copies sold up to that point.
 
  • #9
It should be (Royalties Rate)(Price of Copy)(Number of Copies Sold)
We should have: (0.20)(15)(x-20,000) for x > 20, 000
 
  • #10
Rido12 said:
It should be (Royalties Rate)(Price of Copy)(Number of Copies Sold)
We should have: (0.20)(15)(x-20,000) for x > 20, 000

Which simplifies to 3x-60,000, right?

Thanks.

Edit: And the other two are 1.5x and 2.25x -7,500 when the functions are simplified, right?
 
  • #11
Yep, looks right! (Nod). (If you don't mind me asking, are you learning independently or via course? You don't seem to be in any rush in answering any of your questions.)
 
  • #12
Rido12 said:
Yep, looks right! (Nod). (If you don't mind me asking, are you learning independently or via course? You don't seem to be in any rush in answering any of your questions.)
Awesome, thanks!

It's a self-paced course, so I'm teaching myself. I'll be writing the exam very soon though.
 
  • #13
I'm not too sure if I simplified these correctly. Am I on the right track?

http://i.share.pho.to/c1134604_o.png

Thanks again.
 
  • #14
That looks great! (Yes) I haven't seen you for a while, have you written your exam yet? (Wink)
 
  • #15
Rido12 said:
That looks great! (Yes) I haven't seen you for a while, have you written your exam yet? (Wink)

Yes, I wrote it last week. Thanks to all the pointers you guys gave me with these problem sets, I'm sure I aced it. No doubt about it.

Remember that 2m long string question? Something similar (but not exactly the same question) came up as one of the tough long answer problems and I'm sure I aced it because of you guys. Thank you.
 

FAQ: Piecewise Functions and Domains

What is a piecewise function?

A piecewise function is a function that is defined by different equations or expressions for different intervals of the input, or domain. In other words, the function is divided into different "pieces" that are defined by different rules or equations.

How do you graph a piecewise function?

To graph a piecewise function, you first need to identify the different pieces or intervals of the function. Then, for each interval, you plot the points that satisfy the equation or rule for that interval. Finally, you connect the points to create a piecewise graph.

What is the domain of a piecewise function?

The domain of a piecewise function is the set of all possible input values for which the function is defined. It is important to note that the domain may be different for each piece or interval of the function.

How do you determine the domain of a piecewise function?

To determine the domain of a piecewise function, you need to consider the domain of each individual piece or interval. The domain of the function as a whole will be the set of all input values that are within the domain of at least one of the pieces.

Can the domain of a piecewise function be infinite?

Yes, the domain of a piecewise function can be infinite if one or more of the pieces has an infinite domain. For example, if one piece is defined for all real numbers, then the domain of the function as a whole would also be all real numbers.

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