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Math Question( Pre-calculus)

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Old 02-24-2014, 11:00 AM   #1
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Math Question( Pre-calculus)

I havent really been able to find a way to solve this. I have attempted everything I can think of.

Show the following function is one to one without invoking the horizontal line test.

f(x)= x^2-4x; x<2


P.S- obviously TW is not the place for math advice. Me and my friend have been trying to solve this for two hours. It is a bonus problem and our instructor said we cannot ask him how to do it as it is a bonus, otherwise that is what I would be doing.

Any help would be greatly appreciated!!
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Old 02-24-2014, 11:29 AM   #2
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Its been a while but this is what I got:
y=x^2-4x
complete the square
y+4=x^2-4x+4
y+4=(x-2)^2
(y+4)^(1/2)=x-2
(y+4)^(1/2)-2=x

swap the x and y on the original equation
x=y^2-4y same process and at the end
(x+4)^(1/2)-2=(y+4)^(1/2)-2
which would mean x=y.
so it would be one to one if x<2

I could be way off, that was a while ago.
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Old 02-24-2014, 11:48 AM   #3
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I've been lurking around this forum for a while, thinking about buying a new truck. Anyways, I just was working on my own math for a couple hours and feel your pain so I decided to sign up for tacomaworld and give you a little help.

Hopefully you can read my handwriting, I don't really feel like typing it up. Also, I don't have any experience with what "one to one" means so I am just going off a quick google search. I think this is correct but take a min to make sure it proves what you are looking for.

Good luck.

Edit: It looks like someone beat me to it while I was writing this out.
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Old 02-24-2014, 11:54 AM   #4
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^^^ I should have wrote it out because that is how it should look. I was trying to help it make sense when typing it out.
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Old 02-24-2014, 12:09 PM   #5
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Find the minimum value of the parabola:
f(x)= x^2-4x
0+4=x^2-4x+4
4=(x-2)^2+4
So the minimum value is (2,-4)
Since the minimum occurs at x=2, x less than 2 will always be on the left side of the parabola.
Since the function is constrained to x<2 it is one to one.

Hope this helps. This is kinda just the wordy version of the horizontal line test.

...I used to be good at this stuff.

Both the guys above are right, this is just a different approach.
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Old 02-24-2014, 06:03 PM   #6
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...don't remember what the "horizontal line test" is, or does, ....but you could cheat if you know calculus, ...since the local minima (or maxima, if it exists) can be found by taking the analytical derivative, setting it to zero, solve for x...
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