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I need some help with Calculus *UPDATED QUESTION
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EMP

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PostPosted: Sat Jan 28, 2006 7:10 pm    Post subject:  I need some help with Calculus *UPDATED QUESTION  

how would you take the inverse of
f(x)= 2x^4 - 3x + 5

what is the answer and how did you get it. thanks

EDIT:
what are accumulation functions? (used in calc classes, nothing of that money thing because I tried googling it. )

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OSA


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PostPosted: Sat Jan 28, 2006 11:10 pm    Post subject:  

Isn't the inverse just 1 over all that?

I dunno.

www.math.com

Check there, they might have it.

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krajee

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PostPosted: Sat Jan 28, 2006 11:50 pm    Post subject:  

OSA wrote:
Isn't the inverse just 1 over all that?

Thats reciprocal. Inverse is f^-1(x).

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lld_master

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PostPosted: Sun Jan 29, 2006 3:49 am    Post subject:  

all you have to do is switch the X and the Y and then again make it all equal to Y.

Meaning,

it is:

f(x) = 2x^4 - 3x +5
Y = 2x^4 - 3x +5
X = 2Y^4 - 3Y +5
(X - 5 + 3y) = 2Y^4
(X - 5 + 3y)/2 = Y^4
Y = foruthrootof((X - 5 + 3y)/2)

answer:
y = foruthrootof((X - 5 + 3Y)/2)
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EMP

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PostPosted: Sun Jan 29, 2006 4:42 am    Post subject:  

I can just leave it like that? Don't I have to put all of the 'y's on one side or something? But ((x - 5 + 3y)/2)^(1/4) = y sounds like a plausible answer. The only problem I have with it is that if I were to attempt tio graph it, how would I be able to plug in the y-values if I don't know it?

Thanks

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PostPosted: Sun Jan 29, 2006 12:09 pm    Post subject:  

lld_master wrote:
answer:
y = foruthrootof((X - 5 + 3Y)/2)

y = (x - 5)/(2y^3 - 3) is a bit neater but to each his own.

As for graphing it, it's been awhile since I took Calculus I so I'm not sure on the rules of inverses anymore, sorry.

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LLD_Pie


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PostPosted: Sun Jan 29, 2006 12:58 pm    Post subject:  

to graph it first find its intercepts

so set y to 0 and you get

if y = 0 then x = 5 because 0 / anyting = 0

if y = 1 then

1 = (x-5)/(2-3)
1 = (x-5)/(-1)
-1 = x-5
x = 4

if y = 2 then

2 = (x-5)/(13)
16 = (x-5)
x = 21

if y = -1 then

-1 = (x-5)/(-5)
5 = (x -5)
x = 10

somone check me on this but im pretty sure thats how its done...

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PostPosted: Sun Jan 29, 2006 1:02 pm    Post subject:  

OHHHHHHHHH!! For some reason I thought he wanted it in y= form so that he could plug it into his calculator for an easy quick graph. Oh, okay. Yeah, to graph it by hand (if that's what you want to do), just pick arbitrary y-values and solve for x.
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EMP

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PostPosted: Sun Jan 29, 2006 1:33 pm    Post subject:  

ahh, thanks so much guys.
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PostPosted: Sun Jan 29, 2006 5:11 pm    Post subject:  

god i hated calc =(
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PostPosted: Sun Jan 29, 2006 5:31 pm    Post subject:  

I got kind of dizzy reading this thread.
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PostPosted: Sun Jan 29, 2006 5:32 pm    Post subject:  

Uh oh, i'm taking calc next year... Sad
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PostPosted: Sun Jan 29, 2006 6:06 pm    Post subject:  

to graph this using a calculator would be easy since there is an Inverse function on the TI-83

for example it would be something like this:

y1=Inverse(2x^4 - 3x +5)


done!
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PostPosted: Sun Jan 29, 2006 6:20 pm    Post subject:  

lld_master wrote:
to graph this using a calculator would be easy since there is an Inverse function on the TI-83

The TI-83 only graphs functions, although I think there are programs you can add to change that.

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PostPosted: Sun Jan 29, 2006 11:06 pm    Post subject:  

lld_master wrote:
to graph this using a calculator would be easy since there is an Inverse function on the TI-83

for example it would be something like this:

y1=Inverse(2x^4 - 3x +5)

done!

I whipped out my old TI-83+ and the only thing I found was a InvNorm function. When I plotted it, it gave me a weird scatter of points. I pulled out my TI-89 Titanium (which is what I use for my classes - required) and after another quick search, couldn't even find any sort of Inverse function. Oh well, most of the stuff I do now is either conceptual or requires a program like Mathematica to do.

Gheed wrote:
The TI-83 only graphs functions, although I think there are programs you can add to change that.

Wouldn't doubt it at all.

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