# Rational Function Application

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So I got this HW assignment and i was wondering if anyone could confirm my anwser. Here's the question: The population, in thousands, of Mathville is modeled by the equation P(t) = 20(4t+3)/(2t+5), where t is the time in years since 2005. The question is...what is the population growth rate in 2006. So then I took the formula and found the first derivative, which i found to be 280/(2t+5)^2, and then set the first derivative to equal 0 so i could solve for t. In the end I got 0 = 280. Now i know that doesnt seem right. Can anyone help me?

PS- on the following question i concluded that the population will never reach 50000 because there is a horizontal asymptote at y=40; therefore wont reach 40,000. Correct?

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The question is...what is the population growth rate in 2006. So then I took the formula and found the first derivative, which i found to be 280/(2t+5)^2

Good.

and then set the first derivative to equal 0 so i could solve for t.

Why did you do that? You were asked for the growth rate in 2006, not the when the population becomes stable. (Hint: You know $t$. There is no reason to solve for it.)

With regard to 280=0: You got that by multiplying both sides of $280/(2t+5)^2=0$ by $(2t+5)^2$. The result, 280=0, means there are no finite zeros. The population growth rate curve has zeros at $t=\pm\infty$.

PS- on the following question i concluded that the population will never reach 50000 because there is a horizontal asymptote at y=40; therefore wont reach 40,000. Correct?

Correct, assuming $t$ is non-negative. Negative values of $t$ yield some interesting results. July 2002 ($t = -2.5$) was one heck of a month in Mathville by that population model.

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Good.

Why did you do that? You were asked for the growth rate in 2006, not the when the population becomes stable. (Hint: You know $t$. There is no reason to solve for it.)

With regard to 280=0: You got that by multiplying both sides of $280/(2t+5)^2=0$ by $(2t+5)^2$. The result, 280=0, means there are no finite zeros. The population growth rate curve has zeros at $t=\pm\infty$.

Correct, assuming $t$ is non-negative. Negative values of $t$ yield some interesting results. July 2002 ($t = -2.5$) was one heck of a month in Mathville by that population model.

Thanks a lot! Ok so I get where coming from with setting the first derivative to zero...I missed a few days in class so i was a little confused. I want to find the population growth rate in 2006 so I want to then sub t=6 into the first derivative, which gives me 0.9689 as my population growth rate of 2006 (units?)...is this correct

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I want to find the population growth rate in 2006 so I want to then sub t=6
Why would you do that when:
t is the time in years since 2005

edit, I got the same first derivative.

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Why would you do that when:

edit, I got the same first derivative.

Opppss....thanks a lot! For some reason i thought it was since 2000... Ok so i sub in t=1 into the first derivative and get 40/7 as my population growth rate

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Don't forget to put the units in your answer, that is, thousand people per year.

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Thanks!!! You really helped me out....Its not that I do not get it, it just i'm finding it hard to know when to use like the first derivative oppose to the initial function or when the second derivative is to be properly used.

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Hmm... You sparked my curiosity about july 2002!

$t=-2.5$

$P(t) = \frac{20(4t + 3)}{2t + 5}$

$P(-2.5) = \frac{20(-7)}{0}$

Ooooo... I see, its a asymptote of some kind ...

I assume it goes to negative infinity...

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hmmm....u two seem very intelligent when it comes to calculus so i was wondering if you could confirm this for me...from f(x)= (20x^2-20)/(x^3+x^2) i got f'(x) = (-20x^4 + 60x^2 + 40x)/(x^3 + x^2)^2 can i assume that there are no local extremas? So i now have found f''(x) to be

f''(x)= (-20x^6 + 80x^5 - 60x^4 - 80x^3 - 160x^2)/(x^3 + x^2)^4

How correct am I? I need this info in order to graph the function...i need f''(x) to be correct in order to find concavity and point of inflection. If u could help that would be fantastic

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CanadaAotS do you have any graph plotting software? You should be able to understand the behaviour of the function better if you see it as a whole. It's quite an interesting shape, I'm pretty sure no real-life population models look like that.

anyone interested, in case you didn't know, the rule at play with these derivatives is I think called the quotient rule.

The derivative of $\frac{u}{v}$ wrt x when u and v are functions of x is $\frac{v{\frac{du}{dx}}-u\frac{dv}{dx}}{v^2}$

ender7x77, please, when posting convoluted formulae in this forum, use the inbuilt LaTeX system. Click on any of the images in my post to see the code that I have used.

As for, $f(x)=\frac{20x^{2}-20}{x^{3}+x^{2}}$ I also got $f'(x)=\frac{(-20)x^{4}+60x^{2}+40x}{(x^{3}+x^{2})^2}$, so I guess you're doing OK.

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CanadaAotS do you have any graph plotting software? You should be able to understand the behaviour of the function better if you see it as a whole. It's quite an interesting shape, I'm pretty sure no real-life population models look like that.

anyone interested, in case you didn't know, the rule at play with these derivatives is I think called the quotient rule.

The derivative of $\frac{u}{v}$ wrt x when u and v are functions of x is $\frac{v{\frac{du}{dx}}-u\frac{dv}{dx}}{v^2}$

ender7x77, please, when posting convoluted formulae in this forum, use the inbuilt LaTeX system. Click on any of the images in my post to see the code that I have used.

As for, $f(x)=\frac{20x^{2}-20}{x^{3}+x^{2}}$ I also got $f'(x)=\frac{-20x^{4}+60x^{2}+40x}{(x^{3}+x^{2})^2}$, so I guess you're doing OK.

Thanks....ya i'm new so i never knew how to do the LaTex System. It is called the quotient rule (we just did it two weeks ago). Now am I right to say that there is no local extrema because I cannot solve for x in the first derivative...If so then i have to find the concavity with the second derivative, which i found to be $f''(x)=\frac{-20x^{6}+80x^{5}-60x^{4}-80x^{3}-160x^{2}}{(x^{3}+x^{2})^4}$

I'm sorry if I may be coming off as annoying, but my test a few days ago didnt go as well as i hoped for and now I am feeling very dubious in every mathematical endeavour.

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Assuming that by extrema, you mean stationary point, then there are none of them when there are no values of $x$ for which $f'(x)=0$. The question being, is that the case?[hide]what about when x=2?[/hide]

EvoN1020v, graphical calculators are a tad expensive, don't you think? There is plenty free software for graph plotting which is just as useful if not as portable.

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Assuming that by extrema, you mean stationary point, then there are none of them when there are no values of $x$ for which $f'(x)=0$. The question being, is that the case?[hide]what about when x=2?[/hide]

EvoN1020v, graphical calculators are a tad expensive, don't you think? There is plenty free software for graph plotting which is just as useful if not as portable.

Yeah I'm setting f'(x)=0 to determine whether or not there is a local extrema (local minimum or local maximum), but i do not believe you can solve for x; therefore there is none. So i'm trying to find the concavity (concave up or concave down) in order to get more info on the function in order to graph. EvoN1020v, graphing calculators are a bit expense especially in the case of high school purposes. Also, they can only show me the end result, but i have to be able to explain how i got it. Oh and i downloaded a software that virtually does the same thing...Anyways thanks

Ps~ I've seen the function graphed and there is a turning point, but i found no local extrema...i'm confused

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You're really sure you can't solve for x?

Consider that for $\frac{-20x^{4}+60x^{2}+40x}{(x^{3}+x^{2})^2}$ to be zero, the numerator (top bit) needs to be equal to zero, regardless of what the denominator (bottom bit) is.

Out of interest, what graphing software are you using?

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You're really sure you can't solve for x?

Consider that for $\frac{-20x^{4}+60x^{2}+40x}{(x^{3}+x^{2})^2}$ to be zero, the numerator (top bit) needs to be equal to zero, regardless of what the denominator (bottom bit) is.

Out of interest, what graphing software are you using?

When I set the function to zero and divide out the denominator and factor i get this $-20x(x^{3}-3x-2$ This to me seems unable to be solved...maybe i am just brain dead right now..Oh and the software i am using is FNGraph and i got from downloads.com

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I'm sorry but I honestly can't think of an easy way of solving algebraically.

$0=\frac{-20x^{4}+60x^{2}+40x}{(x^{3}+x^{2})^2}$

$0=-20x^{4}+60x^{2}+40x$

$20x^{4}=60x^{2}+40x$

$x^{3}=3x+2$

$x^{3}-3x=2$

$x(x^{2}-3)=2$

But can you see what value of $x$ fits into there? Or at least that there definitely is one?

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I'm sorry but I honestly can't think of an easy way of solving algebraically.

$0=\frac{-20x^{4}+60x^{2}+40x}{(x^{3}+x^{2})^2}$

$0=-20x^{4}+60x^{2}+40x$

$20x^{4}=60x^{2}+40x$

$x^{3}=3x+2$

$x^{3}-3x=2$

$x(x^{2}-3)=2$

But can you see what value of $x$ fits into there? Or at least that there definitely is one?

ya its x= 2, but i need to be able to solve for x in order to get x=2 which i dont know how to do with what you showed me (i've never seen things done that way)....was my second derivative right?

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I'm pretty sure there isn't a way to solve for x. But so long as you can prove that f(2)=0, then that should be enough.

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I'm pretty sure there isn't a way to solve for x. But so long as you can prove that f(2)=0, then that should be enough.
Ya it makes sense but if i cant explain how to get it then i cant use it...the second derivative seems to be unable to be solved also...I think I'm losing my mind, why would the teacher assign a hw question like this...you can't go from $\frac{x^{2}+x-6}{(x+2)}$ to this. I hate when teachers goes from one side of the spectrum to the other in difficulty.

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$f'(x) = \frac{-20x^4 + 60x^2 + 40x}{(x^3 + x^2)^2}$

So, you got f'(x) and you want f'(x) = 0? This doesn't seem to have any problems lol

$0 = -20x^4 + 60x^2 + 40x$

$0 = -20x(x^3 - 3x - 2)$

For starters there would've been a local extrema at $x = 0$ but $x \not= 0$... you say x = 2 is one, but since this is an equation of degree 4, there will be 4 zero's.

$(-20x)(x^3 - 3x - 2) = 0$

$(-20x)(x - 2)(\frac{x^3 - 3x - 2}{x - 2}) = 0$

$(-20x)(x - 2)(x^2 + 2x - 1) = 0$

$(-20x)(x - 2)(x + \frac{2 \pm \sqrt{8}}{2}) = 0$

$(-20x)(x - 2)(x + 1 \pm \sqrt{2}) = 0$

The roots of f'(x) are $0, 2, 1 + \sqrt{2}, 1 - \sqrt{2}$, and local extrema's are located at all of the roots except 0.

By the way, on step 2 -> 3 I did binomial long division to divide x - 2 into x^3 -3x - 2. I also used the quadratic formula in the last 2 steps.

Hope that helps

EDIT: btw tree, awesome pi site in your sig! lol...

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I'm sorry but I honestly can't think of an easy way of solving algebraically. ...

$x^{3}=3x+2$

FYI, the cubics are soluble by analytical means, proven in the sixteenth century. See http://en.wikipedia.org/wiki/Cubic_equation.

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I know how to solve it. You can use synthetic divsion to figure out the equation that the tree gave you: $x^3=3x+2$

Therefore: $x^3-3x-2=0$. I strongly recommend you to use synthetic division.

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$f'(x) = \frac{-20x^4 + 60x^2 + 40x}{(x^3 + x^2)^2}$

So, you got f'(x) and you want f'(x) = 0? This doesn't seem to have any problems lol

$0 = -20x^4 + 60x^2 + 40x$

$0 = -20x(x^3 - 3x - 2)$

For starters there would've been a local extrema at $x = 0$ but $x \not= 0$... you say x = 2 is one, but since this is an equation of degree 4, there will be 4 zero's.

$(-20x)(x^3 - 3x - 2) = 0$

$(-20x)(x - 2)(\frac{x^3 - 3x - 2}{x - 2}) = 0$

$(-20x)(x - 2)(x^2 + 2x - 1) = 0$

$(-20x)(x - 2)(x + \frac{2 \pm \sqrt{8}}{2}) = 0$

$(-20x)(x - 2)(x + 1 \pm \sqrt{2}) = 0$

The roots of f'(x) are $0, 2, 1 + \sqrt{2}, 1 - \sqrt{2}$, and local extrema's are located at all of the roots except 0.

By the way, on step 2 -> 3 I did binomial long division to divide x - 2 into x^3 -3x - 2. I also used the quadratic formula in the last 2 steps.

Hope that helps

EDIT: btw tree, awesome pi site in your sig! lol...

Thanks! I'm going to look more thoroughly at this tomorrow.

I know how to solve it. You can use synthetic divsion to figure out the equation that the tree gave you: $x^3=3x+2$

Therefore: $x^3-3x-2=0$. I strongly recommend you to use synthetic division.

i don't know what synthetic division...Anyways i'm not sure if i can even present this to my teacher becasue she is going to know that it is above me.

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$f'(x) = \frac{-20x^4 + 60x^2 + 40x}{(x^3 + x^2)^2}$

So, you got f'(x) and you want f'(x) = 0? This doesn't seem to have any problems lol

$0 = -20x^4 + 60x^2 + 40x$

$0 = -20x(x^3 - 3x - 2)$

For starters there would've been a local extrema at $x = 0$ but $x \not= 0$... you say x = 2 is one, but since this is an equation of degree 4, there will be 4 zero's.

$(-20x)(x^3 - 3x - 2) = 0$

$(-20x)(x - 2)(\frac{x^3 - 3x - 2}{x - 2}) = 0$

$(-20x)(x - 2)(x^2 + 2x - 1) = 0$

$(-20x)(x - 2)(x + \frac{2 \pm \sqrt{8}}{2}) = 0$

$(-20x)(x - 2)(x + 1 \pm \sqrt{2}) = 0$

The roots of f'(x) are $0, 2, 1 + \sqrt{2}, 1 - \sqrt{2}$, and local extrema's are located at all of the roots except 0.

By the way, on step 2 -> 3 I did binomial long division to divide x - 2 into x^3 -3x - 2. I also used the quadratic formula in the last 2 steps.

Hope that helps

EDIT: btw tree, awesome pi site in your sig! lol...

Thanks! I'm going to look more thoroughly at this tomorrow.

I know how to solve it. You can use synthetic divsion to figure out the equation that the tree gave you: $x^3=3x+2$

Therefore: $x^3-3x-2=0$. I strongly recommend you to use synthetic division.

i don't know what synthetic division is...Anyways i'm not sure if i can even present this to my teacher becasue she is going to know that it is above me.

OPPPSSS Sorry about the double post my computer lagged.

PS~ If we never have been showed how to solve something like that am I just to assume that it would be acceptable to say that it is unsolvable (word?)

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