Piecewise Functions and Domains

[ardentmed]

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:

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.

 

[Dethrone]

For question 1), instead of defining the interval (1, 2] into 5 different portions, I would define them as one. I.e if $$ 1 < x \le 2$$. 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?

 

[Evgeny.Makarov]

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 $f(x)=4$?

For question 1), instead of defining the interval (1, 2] into 5 different portions, I would define them as one. I.e if $$ 1 < x \le 2$$. 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]

 

[Dethrone]

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.

 

[ardentmed]

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.

 

[ardentmed]

For question 1), instead of defining the interval (1, 2] into 5 different portions, I would define them as one. I.e if $$ 1 < x \le 2$$. 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.

 

[Dethrone]

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 $$(3x + 60,000)$$?

 

[ardentmed]

Can you clarify on how you got $$(3x + 60,000)$$?

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.

 

[Dethrone]

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

 

[ardentmed]

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?

 

[Dethrone]

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.)

 

[ardentmed]

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.

 

[ardentmed]

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.

 

[Dethrone]

That looks great! (Yes) I haven't seen you for a while, have you written your exam yet? (Wink)

 

[ardentmed]

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.