The Furious Angels
FA Discussion => Off Topic => Topic started by: Eroz on January 06, 2005, 05:37:36 pm
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Find the inverse function:
f(x) = 4( x - 2 ) ^ ( 2 / 3 )
I been working on it for a day now and can figure it out.
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I refuse to do this untill school starts again :p
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C'mon man.
Are you seriously asking us to do your homework ?
I could easily do it but I'm not sure you would learn much from it.
What doesn't kill you makes you stronger :)
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I have been trying to do it..... It's just pre-cal. And it not homework it a problem on my last test. (I already had the retake and cant take it again.) It driving me nuts. When I graph it I get half of the graph correct but the other half isn't there and I can't figure out were it is disappearing too.
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math, you cant escape it even in a faction forum.
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hahahahahahaha ....
this is why i'm an english major. i can add, subtract, and balance my checkbook :D
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Here's a hint.
The inverse function has a space somewhere between: "f(x) = -(x ^ ∞)" and "f(x) = x ^ ∞"
:D
Ok I feel generous. Here's what I came up with.
The inverse function of
"f(x) = 4 (x - 2) ^ (2/3)"
is
"f(x)^(-1) = ( (x / 4) ^ (3/2) ) + 2"
It's been over 15 years since I've done this so I'm not 100% sure what an inverse function is supposed to be exactly but it you put "x=10" in the normal function you get "f(x) =16".
If you put "x=16" in the inverse function you get "f(x)^(-1) = 10" i.e. you get the value of the original "f(x)" value. If my memory's good that's how you verify that your calculations hold water.
Another way to check that the inverse function is correct is to replace the value of "x" in the normal function with what you think is the inverse function. Once you isolate "f(x)" you should get something like "f(x) = f(x)" or if you go further "1=1" or even "0=0". If you get that then you know you've got the correct inverse function, assuming you isolated "f(x)" correctly.
Check that out and let me know if you want the intermediate steps.
Dude if you wanna be a good hacker in the Matrix you gotta brush up on those math skills ;)
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I've been out of math for one year. I want to be a comp sci major once I hit up college next year...I'm doomed...
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My pre-calc class was two years ago.... I remember working on those kind of problems but I already forgot how to work them :P
If I learn something I don't apply in life.... within 6 months I will forget it.
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I have a bachelor's degree in Comp Sci. Trust me you need a lot of math.
I'm lucky that God gave me math skills to compensate for my lack of a good singing voice ;)
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Core singing = scarry though.
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At least I can spell "scary" and "thought" ;P
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Ok. I'm not completey nuts, Core. When I did it got that answer and one varation of it. ( f^-1(x) = ((thirdroot(x)^2) / 8 ) +2 ) The problem I am having is when I graph it. For inverse functions should not it look the same but flipped on it said so the x-axis becomes the y-axis But it doesn't it. One half of the graph is missing and I don't like it.
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At least I can spell "scary" and "thought" ;P
yayay lol
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Eroz, I'm taking Pre-Cal too, and I just did that in a test a month ago, and I'm the top math student in my school. So, here you go...
To find the inverse function, substitute the y for the x, then solve for y. In your case it's:
x = 4( y - 2 ) ^ ( 2 / 3 )
x / 4 = ( y - 2 ) ^ ( 2 / 3 )
( x / 4) ^ ( 3 / 2 ) = y - 2
( x / 4 ) ^ ( 3 / 2 ) + 2 = y = f(x)^(-1) or Inverse Function
The graph should look like this:
Blue line - original
Green Line - inverse
(http://img28.exs.cx/img28/9961/erozgraph4rg.jpg)
To check your work, plug the inverse into the original, like this:
f ( f ( x ) ^ -1)
If you have any questions, let me know.
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So we get the same answer Enoc.
I'm happy to know I wasn't way off.
I'm surprised I still knew how to do that after all that time.
:)
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I get that... but when I graph the original function on my calulator (Casio CFX-9850GC (TI 89/82 Eqv)) I get a mirriored part of the graph ( It looks like a absolute function.)
I would show you but I can't find where you made that graph.
CPM - Analysis ????
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If it's in the original function, then it's probably in your placing of parantheses. Those Casio calculators can be tricky. If you don't place them right, it might think you're telling it to graph a quadratic (e.g. x^2), and then to take the third root. On paper, that works, but the calculator needs to graph x to the power of two-thirds (e.g. x^(2/3)). The graph should look like an arm, not a parabola. I don't use those Casios at school so I don't know the exact key layout, or how to work them, so I can't help you there. At school, I use a Texas Instruments TI-83 (they're similar, but not identical) and at home a CPU graphing program.
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No it not quiet the place of my parantheses or not anything I can perse.... if you put it into you TI you should see what I am see (It does the same thing on the 89's as well.) It has something to do with the 2/3 power not being read by the calculator right. Is it a free graphing program because using Excel to graph is annoying.
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Yeah, it's called GraphCalc. I got it off of Downloads.com. Well, the parantheses is the only thing that ever makes my graphs come out wrong on the TI-83 or 89.