anonymous
  • anonymous
A question from my test today...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
can't anyone do it? Where's @TuringTest
anonymous
  • anonymous
can we factorise and cancel to get rid of the pesky denominator? :\[\frac{-x(x-1)(x+3)}{x-1} \le g(x) \le \frac{(x+1)(x-1)}{x-1}\]

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More answers

anonymous
  • anonymous
i guess, i didnt have time to do it, Last question of the test... ;/
anonymous
  • anonymous
so: \[-x(x+3) \le g(x) \le (x+1)\] and you want the limit as x tends to 1 ?
anonymous
  • anonymous
yup. :)
anonymous
  • anonymous
well evaluating at x = 1 we have \[-4 \le g(x) \le 2\] im not sure whether thats what you want or not, i haven't really seen this kinda Q before
anonymous
  • anonymous
it isn't, there's only one value. ;/ But thank you :))
anonymous
  • anonymous
hmm
anonymous
  • anonymous
call in the cavalry! @FoolForMath
anonymous
  • anonymous
My mind is officially fried for the day :( sorry.
anonymous
  • anonymous
aw ok
anonymous
  • anonymous
@TuringTest
TuringTest
  • TuringTest
whoa what happened? let em read this
anonymous
  • anonymous
haha
TuringTest
  • TuringTest
the limit as x goes to 1 ?
zepp
  • zepp
Apply the squeeze theorem to the question :)
TuringTest
  • TuringTest
what limit is this, I don't understand the notation
anonymous
  • anonymous
\[\lim_{x \rightarrow 1}\]
experimentX
  • experimentX
looks like 2
TuringTest
  • TuringTest
I hope smoothmath's cookin up something good 'cuz I see what Eigen saw... how do you figure X ?
anonymous
  • anonymous
i factorised wrong i think
TuringTest
  • TuringTest
oh I trusted you! ok I'll check
anonymous
  • anonymous
derp
anonymous
  • anonymous
sorry guys
experimentX
  • experimentX
-(x-1)(x-3)
anonymous
  • anonymous
yah. sorry im exhausted, just finished ~ 4hrs physics work
anonymous
  • anonymous
I like l'hopitals rule for both of those equations. To use L'hopital's rule, we have to make sure they satisfy the condition that the functions at that point are of the indeterminate form 0/0 or inf/inf. These both look like 0/0, so we're good to go. Now, since we've satisfied that condition, L'hopital's rule says that the limit of each function will be the same as if we take the derivative of the top and bottom. So basically, if our function is f(x)/g(x), the rule says that the limit will be the same as f'(x)/g'(x), as long as that limit exists. What that gives us is: (-3x^2 + 8x - 3)/1 <= g(x) <= 2x/1 Evaluating at x=1 gives: (-3 + 8 -3)/1 <= g(x) <= 2/1 2 <=g(x) <=2 So, by squeeze theorem, lim g(x) is 2.
anonymous
  • anonymous
in fact ima go sleep night!
anonymous
  • anonymous
Haha don't worry, Turing. I cooked up something delicious.
TuringTest
  • TuringTest
I can see that! that's what I get for not double-checking other peoples algebra...
experimentX
  • experimentX
well that was delicious.
anonymous
  • anonymous
good night, and thank you :P We can't use L'hopital's rule..he said he wouldn't consider the answer ;/
anonymous
  • anonymous
Ugh. Sucks. I hate that he disallowed it because it works so well...
anonymous
  • anonymous
i knoow.
experimentX
  • experimentX
factorize and cancel ... that's quite easier that L'hospital's rule.
TuringTest
  • TuringTest
but it does factor to\[{-x(x-1)(x-3)\over x-1}\le g(x)\le x+1\]\[-x(x-3)\le g(x)\le x+1\]so yeah, we can just plug in like experimentX says
experimentX
  • experimentX
well ... still, I really liked that squeezing theorem.
anonymous
  • anonymous
Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.
experimentX
  • experimentX
real mathematicians do that ..!!!
anonymous
  • anonymous
But yeah, to solve by factoring, turing's method above is correct. Eigen's original misfactoring is what gave the error we got.
anonymous
  • anonymous
Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.
experimentX
  • experimentX
you are indeed a real one!!!
TuringTest
  • TuringTest
agreed :)
anonymous
  • anonymous
Haha okay. Cool. Both methods are great, and I actually think factoring is easier to teach. I just find L'hopital's easier to execute. And sexier.
TuringTest
  • TuringTest
haha you should be "SexyMath" then
anonymous
  • anonymous
omg
experimentX
  • experimentX
LOL..
anonymous
  • anonymous
thanks :)
anonymous
  • anonymous
lols. That might give the wrong impression, but I like it.

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