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Author Topic: Mathematics Help Thread  (Read 228798 times)

Descan

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Re: Mathematics Help Thread
« Reply #1260 on: October 04, 2013, 10:21:18 pm »

There is no other function. Those are the only two things given to me.

The question asked is:

Describe the transformation that must be applied to the graph of the base function f(x) to obtain the transforming function. Write the transformed equation in simplified form.

f(x) = x^2, y = 7f(-1/6(x-1))+1

There is no other function, equation, or anything. Nothing to plug it into.

And on that note, wouldn't "Plugging -1/6(x-1) into the function f, then multiplying by 7 and adding 1,", if that means "Replace f(x) = x^2 with f(x) = (-1/6(x-1)^2, then multiply by 7 and add 1", basically result in an equation that looks like y = 7f(-1/6(x-1))^2+1? Or possibly y = 7f((-1/6(x-1))^2)+1. *

And finally, as I said, it's transforming the graphed f(x) = x^2. Nothing algebraic there, as far as I can tell.

I'm not saying you're wrong. These are my assumptions, they're fully capable of being wrong. But they're the assumptions I got from the rest of the book, so if they're wrong I'm either a dumbass or the book is terrible at teaching. Or both :v

*I ran with it and simplified y = 7f((-1/6(x-1))^2)+1, and the result is exactly the same as simplifying y = 7f(-1/6(x-1))^2+1 (the equation I assumed the book -meant- to write and forgot the exponent to), that is to say, y = (-7(x-1/6)^2)+1
« Last Edit: October 04, 2013, 10:28:39 pm by Descan »
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Re: Mathematics Help Thread
« Reply #1261 on: October 04, 2013, 10:25:40 pm »

Yes. Sorry, I meant you were wrong about them writing it down wrong, that is exactly what you're meant to do. Only remove the f from the equation as it's now superfluous.
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Descan

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Re: Mathematics Help Thread
« Reply #1262 on: October 04, 2013, 10:30:50 pm »

Okay. I'll keep that in mind. When it has an f in the y = equation, even if the y = equation doesn't look right as compared to the f(x), stick the parts in brackets after f into the original f(x) = (whatever) as fits.
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Re: Mathematics Help Thread
« Reply #1263 on: October 04, 2013, 10:42:38 pm »

I think this sort of thing tends to confuse people because you're dealing with two different versions of the variable x.

You've got one version in the function f(x)= x2. Then you've got another in the equation y = 7f(-1/6(x-1)) + 1. Even though they're pretty much unrelated.

Just keep in mind the f(x)= x2 is just telling you what the function does. The equation is actually asking you to use it to create a graph.
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Descan

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Re: Mathematics Help Thread
« Reply #1264 on: October 04, 2013, 10:50:05 pm »

The problems I've found with this book all basically boil down to "Typoes" and "Unclear instructions"

Sometimes the examples will be poor, or the instructions will teach you one way to do this that don't actually work in all cases, and then ask you to answer a case where the taught method doesn't actually work.

And sometimes it'll give you a crutch, in this case by previously sticking the ^2 or the 1/x or the sqrt x or whatever inside the y = equation, instead of having you re-jigger the equation via the f(x) stuff we just went over. And then it'll screw you over by not telling you that it IS a crutch, and not telling you when it's going to take that crutch away.
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da_nang

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Re: Mathematics Help Thread
« Reply #1265 on: October 05, 2013, 12:59:10 am »

The problems I've found with this book all basically boil down to "Typoes" and "Unclear instructions"

Sometimes the examples will be poor, or the instructions will teach you one way to do this that don't actually work in all cases, and then ask you to answer a case where the taught method doesn't actually work.

And sometimes it'll give you a crutch, in this case by previously sticking the ^2 or the 1/x or the sqrt x or whatever inside the y = equation, instead of having you re-jigger the equation via the f(x) stuff we just went over. And then it'll screw you over by not telling you that it IS a crutch, and not telling you when it's going to take that crutch away.
One of the reasons why word problems are sometimes the bane of mathematics education.

Also, nitpicking word from the wise: a/bc is an ambiguous notation on the internet.
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1266 on: October 05, 2013, 01:17:07 am »

For a piecewise function, if (1) the slope at the boundary between each piece was the same, (2) the value at the boundary was the same, would it be smooth and differentiable over its entire domain?

For example:

If (x>= 1) y = x^2+3x-1
if (x < 1) y = x^3+2x

when x=1 both parts have the same slope and value.

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Sixteen

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Re: Mathematics Help Thread
« Reply #1267 on: October 05, 2013, 01:26:52 am »

I have a problem, but it's mostly because I have no clue what the wording of a problem means. The problem asks to find the "hyperbola associated to the logarithm log10(x)," but I've never run into this wording before and the only google result I get is the document it comes from.
I've talked to my professor and he says he covered it last lecture, but looking over my notes for that day, I'm pretty sure he didn't use those terms. The only thing I could find that was related would be the conversion to ∫(1/x)dx form, in which case the answer would be ∫1/(x*ln(10))dx. Does this sound right to you guys, or do you know what that phrasing means?
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da_nang

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Re: Mathematics Help Thread
« Reply #1268 on: October 05, 2013, 02:10:53 am »

For a piecewise function, if (1) the slope at the boundary between each piece was the same, (2) the value at the boundary was the same, would it be smooth and differentiable over its entire domain?

For example:

If (x>= 1) y = x^2+3x-1
if (x < 1) y = x^3+2x

when x=1 both parts have the same slope and value.
Assuming each function part is differentiable over their specific domains, then the piecewise function would be differentiable. Smooth, not necessarily.
E.g.
y1 = 0, x<0
y2 = x2, x>=0

This piecewise function is differentiable, but not smooth as it has no second derivative at x=0 since y'1 = 0 and y'2 = 2x which clearly have different slopes at x=0.

I have a problem, but it's mostly because I have no clue what the wording of a problem means. The problem asks to find the "hyperbola associated to the logarithm log10(x)," but I've never run into this wording before and the only google result I get is the document it comes from.
I've talked to my professor and he says he covered it last lecture, but looking over my notes for that day, I'm pretty sure he didn't use those terms. The only thing I could find that was related would be the conversion to ∫(1/x)dx form, in which case the answer would be ∫1/(x*ln(10))dx. Does this sound right to you guys, or do you know what that phrasing means?
"Association" is a rather vague and ambiguous term in that problem. Your suggestion is a possibility but without further context/information it's entirely speculative.
« Last Edit: October 05, 2013, 02:30:13 am by da_nang »
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Re: Mathematics Help Thread
« Reply #1269 on: October 05, 2013, 02:40:48 am »

Oh well, thanks anyways, I'll take what I can get. If worst comes to worst and it's actually used in a graded assignment, I'll just throw whatever I can think of at it and hope it scores some points.
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1270 on: October 05, 2013, 02:54:09 am »

Don't y'=0 and y'=2x obviously have the same slope at x=0? ohwait stupid me
« Last Edit: October 05, 2013, 02:56:49 am by Skyrunner »
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1271 on: October 05, 2013, 08:38:34 am »

Transform f(x) = x^2 as a graph into y = 7f(-1/6(x-1))+1

First of all, this question is completely valid and correct, and second of all, there is actually a graph transformation that transforms graph(f) into graph(y=7f(-1/6(x-1))+1), regardless of what f actually is.

In fact, you can read off the transformation by looking at your second equation. The transformation looks like this:
  • Move the graph one to the right:   (h(x) -> h(x-1))
  • Stretch the graph along the x axis by the factor -6:   (h(x) -> h(-1/6*x))
  • Stretch the graph along the y axis by the factor 7:   (h(x) -> 7*h(x))
  • Move the graph one up:   (h(x) -> h(x)+1)

I'm pretty sure that those four kinds of elementary transformations are all you actually need for your class, and any such graph transformation problem can be solved by combining those transformations in the correct order.
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1272 on: October 05, 2013, 11:37:34 am »



Maybe you should use a Fourier transformation.
 They seem to solve everything.


(jk)
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Descan

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Re: Mathematics Help Thread
« Reply #1273 on: October 05, 2013, 02:10:32 pm »

Magma, we already solved that issue. And that part was obvious so I didn't feel the need to ask how to do it. Already had all that written down. :v
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Descan

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Re: Mathematics Help Thread
« Reply #1274 on: October 06, 2013, 09:21:02 pm »

Fucking Math, how does it work?

"A jet flew from Tokyo to Bangkok, a distance of 4800 km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flights was 2 hours, what was the jet's speed from Bangkok to Tokyo?"

I've gone through about 5 different pages and I always end up with some monstrosity with 480k or 960k = y and I'm pretty sure it doesn't take 960 000 hours to fly to Tokyo :U

So I have no idea what the variables should be or how to arrange the equation. :I

I -thought- it was 4800/x + 4800/(x-200) = 2y+2

With x being speed, and y being the time it takes to make one trip from Tokyo to Bangkok.

Edit: Got 4800x - 239 999 = y. Lesse what I can do with that...

Edit: Okay pretty sure it's not supposed to give me an x^3. :I
« Last Edit: October 06, 2013, 09:51:42 pm by Descan »
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