Practical domain

[renegade05]

On this question:

\(\displaystyle T(m)=\frac{17-2m}{7m-162}\)

Where m = market share and t = number months since the product was placed on the market.

I graphed this on my graphing calculator and it looks like it has 2 X intercepts.

The question is find the practical domain of the equation. So, i assumed it would be from the 2 X intercepts since the output(time) cannot be negative.

But of course, looking at this function you can tell right away there is only going to be one answer when plugging in 0 for m.

First question, why does it look like it has 2 x intercepts on the graphing calculator?
Second question, how to find the practical domain. Is this where approching limits comes into play? Something i have not learned yet.

Thanks!

 

[mmm4444bot]

\(\displaystyle T(m)=\frac{17-2m}{7m-162}\)

Where m = market share and t = number months since the product was placed on the market.

Watch out. The symbols t and T do not have the same meaning; they are not interchangable.

T = number of months since product debut

OR

t(m) = (17 - 2m)/(7m - 162)

This function seems like an inverse, to me.

I mean, time is not a function of market share; it's the other way around, yes?

Otherwise, time (reality) would be forced to change as companies' market share fluctuates. If one company's m goes up and another company's m goes down, how would T know what to do? Would two months suddenly turn into one month or three?

The posted description of T(m) does not make sense, to me.



I graphed this on my graphing calculator and it looks like it has 2 X intercepts.

The graph of the following equation has one x-intercept.

y = (17 - 2x)/(7x - 162)



First question, why does it look like it has 2 x intercepts on the graphing calculator?

You might need to adjust the domain and range on the machine window, to get a better view of what y is doing and where.



Second question, how to find the practical domain.

Well, we would expect market share to go from 0% through 100%, to cover all the possibilities, yes?

But, the given function does not make sense over that domain for m because, as you pointed out, a negative number of months does not make sense.

Therefore, if the posted function is correct, then that mathematical model only makes sense for a restricted span of percents. So, you need to look at good graphs, to see where the graph makes sense and where it does not.

That's what they mean by "practical".

Do you know how to use algebra, to solve T(m) = 0 for m, to confirm the exact value of the intercept that you see on the graph ?

Also, I'm curious to know, is your teacher a machine ?

The following images show the behavior of T(m) with both a large domain (global view) and a domain restricted to the "practical" area.

(Double-click images to expand, if needed.)

Cheers ~ Mark

[attachment=1:2xwhmy71]Global view.JPG[/attachment:2xwhmy71]

[attachment=0:2xwhmy71]Restricted view.JPG[/attachment:2xwhmy71]