I will if I get stuck. I don't want to just get the answers without attempting them. I won't succeed that way. Thanks though.

2/9/2010 2:58:43 PM

such honesty

2/9/2010 3:00:09 PM

Let f(x) = x^3.

Estimate the values of the following by using a graphing device to zoom in on the graph of f. Give your answers correct to one decimal place.

I am banging my head against the wall here when doing f'(2) and f'(3). I've graphed them on Wolframalpha to ridiculously small lengths, and it's still wrong.

For example, on f'(3): [link]http://www.wolframalpha.com/input/?i=Plot[x^3,+{x,+2.999,+3.001}][/link]

That LOOKS like (27.01 - 26.99) / (3.0003 - 2.99962), but that's WRONG.

And on f'(2): [link]http://www.wolframalpha.com/input/?i=Plot[x^3,+{x,+1.99,+2.01}][/link]

That LOOKS like (8.00 - 7.95) / (2 - 1.996), but that is also wrong.

I'm pretty sure the problem is that Webassign is looking for an incredibly specific value, my eyeballing skills just aren't perfect. I've already used like 5 submissions on these two.

[Edited on February 9, 2010 at 3:49 PM. Reason : ]

2/9/2010 3:48:50 PM

so why not just find it exactly?

take the derivative and plug in the value

[Edited on February 9, 2010 at 3:51 PM. Reason : 12 and 27 i think]

2/9/2010 3:50:18 PM

I tried that, and I end up with

lim h->0 of

(2x^3 + 3hx^2 + 3xh^2 + h^3) / h

And I can't figure out how to simplify that.

[Edited on February 9, 2010 at 4:00 PM. Reason : ]

2/9/2010 3:57:14 PM

f(x)=x^3
f'(x)=3x^2

2/9/2010 3:59:11 PM

Yes I know that (via Wolfram Alpha) but I don't know how they derived 3x^2

I've simplified the equation down to:

lim h->0

(2x^3)/h + 3x^2

2/9/2010 4:00:27 PM

that's just the shortcut way. just plug in the value.

f(x)=x^a
f'(x)=ax^(a-1)

always

[Edited on February 9, 2010 at 4:02 PM. Reason : afds]

2/9/2010 4:01:45 PM

"(2x^3 + 3hx^2 + 3xh^2 + h^3) / h

And I can't figure out how to simplify that."

Somewhere in your work you got a sign wrong for one of your x^3 terms. They should've canceled, leaving:

(3hx^2 + 3xh^2 + h^3) / h

instead.

2/9/2010 4:05:14 PM

THis is one of the sadder threads i've see on TWW>

2/9/2010 4:09:22 PM

^^Oops, it was a minus instead of a plus

2/9/2010 4:11:11 PM

have they changed the name from "calculus" to "calculator tricks" yet

2/9/2010 4:11:23 PM

you're zoomed in too close, when you are zoomed in that close your error is multiplied (dividing by 3.003-2.99962 is the same as multiplying by 1,470):

[link]http://www.wolframalpha.com/input/?i=Plot[x^3,+{x,+2.900,+3.100}][/link]

(28.5-25.8)/(3.05-2.95)=27

2/9/2010 4:12:49 PM

9 posts later and it's still 27

2/9/2010 4:15:16 PM

lulz

2/9/2010 4:18:20 PM

I don't get why everyone finds this so amusing, or sad.

[Edited on February 9, 2010 at 4:19 PM. Reason : ]

2/9/2010 4:18:55 PM

Quote :

"Yes I know that (via Wolfram Alpha) but I don't know how they derived 3x^2 I've simplified the equation down to: lim h->0 (2x^3)/h + 3x^2"

the value of

lim h->0

(2x^3)/h

is 0, and since there is no "h" in the second term ( 3x^2), it's left standing, and is therefore your answer.

2/9/2010 4:20:10 PM

^that is incorrect

2/9/2010 4:21:33 PM

It's correct enough

2/9/2010 4:27:02 PM

not in my opinion

[Edited on February 9, 2010 at 4:29 PM. Reason : that limit goes to infinity]

2/9/2010 4:28:32 PM

The population P (in thousands) of Belgium from 1992 to 2000 is shown in the table. (Midyear estimates are given.)



Estimate the instantaneous rate of growth in 1996 by measuring the slope of the tangent.

I'm not sure how to do this. Hint?

2/9/2010 4:29:06 PM

slope=rise/run=(10.175-10.152)/(1998-1996)

[Edited on February 9, 2010 at 4:31 PM. Reason : units would the thousands of people per year]

2/9/2010 4:30:51 PM

Why 1998 to 1996?

That came out to 11.5, which was wrong.

[Edited on February 9, 2010 at 4:32 PM. Reason : ]

2/9/2010 4:31:22 PM

you either have to go with 96 to 98 or 94 to 96. they probably wrote the question to give you similar results both ways. check them both if you want.

[Edited on February 9, 2010 at 4:33 PM. Reason : or you could do 94 to 98. not sure]

2/9/2010 4:32:31 PM

I did. They're both wrong.

I've also tried 1998 to 1994, it's wrong as well.

[Edited on February 9, 2010 at 4:33 PM. Reason : ]

2/9/2010 4:33:04 PM

so does it want the answer in people/year?

[Edited on February 9, 2010 at 4:35 PM. Reason : you could graph all those points, fit a curve, then take the derivative]

[Edited on February 9, 2010 at 4:36 PM. Reason : wait, are those commas or decimals?]

2/9/2010 4:34:32 PM

Thousand people per year

2/9/2010 4:35:39 PM

add both 94-96 and 96-98 and divide by 4

[Edited on February 9, 2010 at 4:56 PM. Reason : ]

2/9/2010 4:35:42 PM

^that's the same as doing 94 to 98

2/9/2010 4:37:27 PM

You'd probably maybe want 1998 to 1994...

maybe they want you to use excel to calculate the linear regression, then find the derivative of that curve.

[Edited on February 9, 2010 at 4:39 PM. Reason : ]

2/9/2010 4:38:09 PM

It's population P in thousands. So 10,196 people, for example.

2/9/2010 4:38:41 PM

how the hell is it a thousand per year?

2/9/2010 4:39:49 PM

wait, is this Chit Chat?

2/9/2010 4:40:01 PM

^^ 11.5 would mean "11.5 thousand people per year" or 11,500 people per year.

2/9/2010 4:41:15 PM

ooh, gotcha. hold on then

2/9/2010 4:41:44 PM

what'd you get for 96 to 94?

2/9/2010 4:44:43 PM

21.5

edit: (10152 - 10109) / (1996 - 1994)

edit: which is wrong, by the way.

[Edited on February 9, 2010 at 4:48 PM. Reason : ]

2/9/2010 4:46:35 PM

try 18.3

2/9/2010 4:53:35 PM

have you learned linear regression?

2/9/2010 4:53:45 PM

The change between 94-96 is 21.5/year, and 96-98 is 11.5/year. Average those for is 16.5/year. I dunno what answers you've tried yet.

2/9/2010 4:55:17 PM

Quote :

"try 18.3"

Nope, wrong.

Quote :

" have you learned linear regression?"

Not yet, just basic derivatives.


^And I've tried 16.5, it's also wrong.

[Edited on February 9, 2010 at 4:56 PM. Reason : ]

2/9/2010 4:56:02 PM

i fit a quadratic equation to the data and it had a slope of about 18.3 at 1996

[Edited on February 9, 2010 at 4:57 PM. Reason : that's pretty much the most accurate way to do this, so i'm not sure wtf they want]

2/9/2010 4:56:22 PM

Well then webassign is lying, lol.

2/9/2010 4:56:54 PM

it should be 16

[Edited on February 9, 2010 at 4:58 PM. Reason : (10175-10109)/4]

2/9/2010 4:57:27 PM

^that's 16.5, which he said is wrong

2/9/2010 5:00:14 PM

Quote :

"I don't get why everyone finds this so amusing, or sad."

rly?

2/9/2010 5:00:47 PM

this thread is getting ridiculous.

I say there should be bare titties on every page, or it should be locked/moved.

2/9/2010 5:02:03 PM

That population problem (as you've stated it at least) is bullshit. "Instantaneous rate of change" of a function has no meaning if the domain is a discrete subset of the reals.

Is there more to the problem that you've left out? Are you supposed to be using a particular population model, such as exponential or logistic? If not, I would complain to the professor; no math professor that I'm aware of would consider that to be a well-written problem.

2/9/2010 5:59:32 PM

This isn't one of those situations where webassign is being douchey about significant digits is it, ie 21.5 should be correct, but does it want 21.500 or something?

2/9/2010 6:24:32 PM

Quote :

"no math professor that I'm aware of would consider that to be a well-written problem writes his own webassign problems."

I'm just sayin'

2/9/2010 7:05:43 PM