A question from my test today...

- anonymous

A question from my test today...

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

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- anonymous

can't anyone do it? Where's @TuringTest

- 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

i guess, i didnt have time to do it, Last question of the test... ;/

- anonymous

so:
\[-x(x+3) \le g(x) \le (x+1)\]
and you want the limit as x tends to 1 ?

- anonymous

yup. :)

- 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

it isn't, there's only one value. ;/ But thank you :))

- anonymous

hmm

- anonymous

call in the cavalry! @FoolForMath

- anonymous

My mind is officially fried for the day :( sorry.

- anonymous

aw ok

- anonymous

@TuringTest

- TuringTest

whoa what happened?
let em read this

- anonymous

haha

- TuringTest

the limit as x goes to 1 ?

- zepp

Apply the squeeze theorem to the question :)

- TuringTest

what limit is this, I don't understand the notation

- anonymous

\[\lim_{x \rightarrow 1}\]

- experimentX

looks like 2

- TuringTest

I hope smoothmath's cookin up something good 'cuz I see what Eigen saw...
how do you figure X ?

- anonymous

i factorised wrong i think

- TuringTest

oh I trusted you!
ok I'll check

- anonymous

derp

- anonymous

sorry guys

- experimentX

-(x-1)(x-3)

- anonymous

yah. sorry im exhausted, just finished ~ 4hrs physics work

- 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

in fact ima go sleep night!

- anonymous

Haha don't worry, Turing. I cooked up something delicious.

- TuringTest

I can see that!
that's what I get for not double-checking other peoples algebra...

- experimentX

well that was delicious.

- anonymous

good night, and thank you :P We can't use L'hopital's rule..he said he wouldn't consider the answer ;/

- anonymous

Ugh. Sucks. I hate that he disallowed it because it works so well...

- anonymous

i knoow.

- experimentX

factorize and cancel ... that's quite easier that L'hospital's rule.

- 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

well ... still, I really liked that squeezing theorem.

- anonymous

Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.

- experimentX

real mathematicians do that ..!!!

- 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

Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.

- experimentX

you are indeed a real one!!!

- TuringTest

agreed :)

- 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

haha you should be "SexyMath" then

- anonymous

omg

- experimentX

LOL..

- anonymous

thanks :)

- anonymous

lols. That might give the wrong impression, but I like it.

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