anonymous
  • anonymous
helpppp: find the critical points: s(t)= (t-1)^4 (t+5)^3
Calculus1
schrodinger
  • schrodinger
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anonymous
  • anonymous
@mathcalculus , what have you tried?
anonymous
  • anonymous
yes, i used the product rule: \[\frac{ d }{ dx } (t-1)^{4}* (t+5)^{3}= (t-1)^{4} * 3(t+5)^{2} + (t+5)^{3} * 4(t-1)^{3}\]
anonymous
  • anonymous
then from there, i'm stuck.

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anonymous
  • anonymous
ok, so far so good. Now, factor out the greatest common factor of both terms. For example, they both have a factor of (t-1)^3, but that's not all they have in common. What else?
anonymous
  • anonymous
?
anonymous
  • anonymous
Don't they both have a factor of t+5 to a power?
anonymous
  • anonymous
oh yes, but are from opposite sides..
anonymous
  • anonymous
doesn't matter the order. AB+CA=A(B+C)
anonymous
  • anonymous
can you explain to me. i dont really see where we are going with factoring
anonymous
  • anonymous
The purpose in factoring is to simplify the derivative into a product of three factors. One of the factors will be t-1 to a power, a second factor will be t+5 to a power, and then you will be left with some factor containing a t as well.
anonymous
  • anonymous
And then once you have your derivative in factored form, you can set each factor equal to zero to find your critical points.
anonymous
  • anonymous
can you show me
anonymous
  • anonymous
i understand what youre saying but not how sure how to do that.
anonymous
  • anonymous
Ok, lets look at the t-1 term in each expression.
anonymous
  • anonymous
ok
anonymous
  • anonymous
What is the largest exponent they share?
anonymous
  • anonymous
3
anonymous
  • anonymous
now i know that i write (t-1)^3
anonymous
  • anonymous
but what happens to the 4?
anonymous
  • anonymous
Exactly, so, we may factor out a \[\left( t-1 \right)^{3}\] from both expressions. Hang on, we will come to that.
anonymous
  • anonymous
k
anonymous
  • anonymous
Let's take a looke at the first expression in its entirety. If all you are removing is the \[\left( t-1 \right)^{3}\] You would be left with \[4\left( t+5 \right)^{3}\]
anonymous
  • anonymous
Do you see that?
anonymous
  • anonymous
no if i was to see the common factor.. i would get (t-1)^3+ (t+5)^2.... then what do i do with the numbers that are left...?
anonymous
  • anonymous
ok,, there is your mistake. It isn't a sum. It's a product.
anonymous
  • anonymous
ok so i dont see how you got 4(t+5)^3
anonymous
  • anonymous
Hang on, I was only talking about factoring out the t-1 from the first expression,, not from the whole thing.
anonymous
  • anonymous
well you showed me that
anonymous
  • anonymous
Hang on my pc is giving me fits.
anonymous
  • anonymous
kk
anonymous
  • anonymous
|dw:1363286087022:dw|
anonymous
  • anonymous
why 4(t+5)?????
anonymous
  • anonymous
On the outside is what they both have in common, on ther inside is what is left after you have factored out that common factor.
anonymous
  • anonymous
i understand this part... (t-1)^3* (t+5)^2
anonymous
  • anonymous
Rememmber when I was saying if all we factored out was the (t-1)^2 You would have the 4(t+5)^3 left.
anonymous
  • anonymous
but then i dont understand why it is 4(t+5) ... i did this..4(t-1) + 3(t+5)
anonymous
  • anonymous
Well, now we include a (t+5)^2 as well, so what is left from the 4(t+5)^3? Just the 4(t+5).
anonymous
  • anonymous
okkk
anonymous
  • anonymous
i understand...
anonymous
  • anonymous
im left with 7t+17
anonymous
  • anonymous
t= -17/7
anonymous
  • anonymous
Perfect. So, now that each expression of the derivative is writtern in factored form, we set each of them equal to zero to find your critical points.
anonymous
  • anonymous
That is one of your critical points. There are two others.
anonymous
  • anonymous
ok so (t-1)^3 = 0
anonymous
  • anonymous
Good.
anonymous
  • anonymous
x= 0?
anonymous
  • anonymous
Not quite.
anonymous
  • anonymous
If (t-1)^3=0 then t=1
anonymous
  • anonymous
Now you have one more to do.
anonymous
  • anonymous
1? how?
anonymous
  • anonymous
Tinme for lunch, I will come back shortly to see if you have finished it.
anonymous
  • anonymous
the exponent is in the way.
anonymous
  • anonymous
Well, raise each side to the 1/3 power. That will get rid of your exponent.
anonymous
  • anonymous
brb
anonymous
  • anonymous
kkk but how do we find the y
anonymous
  • anonymous
Just go back in and substitute the critical values.
anonymous
  • anonymous
For t=1 and t=-5, you will get y=0. But for t=-17/7, you won't get y=0.
anonymous
  • anonymous
i did....
anonymous
  • anonymous
(1-1)^4 (1+5)^3
anonymous
  • anonymous
=0
anonymous
  • anonymous
and -17/7 i got a huge number.. doesn;t make sense.. :(
anonymous
  • anonymous
i got -432/823543
anonymous
  • anonymous
Well, you will need to raise some fractions with a denominator of 7 to the 4th and 3rd powers, so yeah, it will get rather ugly!
anonymous
  • anonymous
??
anonymous
  • anonymous
i see that but it keeps saying the critical points are wrong.
anonymous
  • anonymous
=(
anonymous
  • anonymous
ok, I need to see the original problem. And what keeps saying your critical points are wrong?
anonymous
  • anonymous
Are you using an online program?
anonymous
  • anonymous
webwork.
anonymous
  • anonymous
yes
anonymous
  • anonymous
hang on.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
i attached it.
anonymous
  • anonymous
Your value for t=-17/7 is incorrect
anonymous
  • anonymous
why?
anonymous
  • anonymous
You should get a large numerator divided by a denominator of 7^7.
anonymous
  • anonymous
so i dont know where i went wrong...
anonymous
  • anonymous
The numerator should be (-24)^4(18)^3
anonymous
  • anonymous
we did this step by step... now t is wrong?? really?
anonymous
  • anonymous
No, t is not wrong, but your evaluation of the function for t=-17/7 is incorrect.
anonymous
  • anonymous
so now how do i solve that?
anonymous
  • anonymous
obviously i know its wrong.
anonymous
  • anonymous
ive tried this problem out 3294810347 times.
anonymous
  • anonymous
now what do i do?
anonymous
  • anonymous
Should have (-17/7-1)^4(-17/7+5)^3
anonymous
  • anonymous
huhhh???
anonymous
  • anonymous
\[-\frac{ 17 }{ 7 }-1=-\frac{ 24 }{ 7 }\] and \[-\frac{ 17 }{ 7 }+5=\frac{ 18 }{ 7 }\]
anonymous
  • anonymous
wow this is insane. ive spent an hour to 2 with this problem.
anonymous
  • anonymous
why?? why did we ignore the exponents???
anonymous
  • anonymous
Now raise each of those terms to the appropriate powers.
anonymous
  • anonymous
sn;t it suppose to be part of it?? the original problem we were given??
anonymous
  • anonymous
i did
anonymous
  • anonymous
we didn't I was just showing you step by step.
anonymous
  • anonymous
thats how i got the hugeeeeeeee number
anonymous
  • anonymous
and so did i. i had it on paper.
anonymous
  • anonymous
what i dont understand is that youre basically saying raise those to the power.. in other words, im redoing this whole crap again.
anonymous
  • anonymous
Well, the numerator should be larger than your denominator, and it wasn't according to your answer.
anonymous
  • anonymous
which AGAIN, is leading m to the wrong huge answer i had before
anonymous
  • anonymous
No, you are finding the y-value that goes with the value you ccame out for t.
anonymous
  • anonymous
so there are 4 crtical points now?
anonymous
  • anonymous
or 3??
anonymous
  • anonymous
youre saying raise hem to the power and multiple right???
anonymous
  • anonymous
do you know rasing them to the power is going to GIVE ME A HUGEEEEEE NUMBER??
anonymous
  • anonymous
do you not see that was what i had in the first place??
anonymous
  • anonymous
There are only the three values of t we came up with, but you want them as ordered pairs, yo
anonymous
  • anonymous
ok good3 values done.
anonymous
  • anonymous
obviously i want them in order pairs.
anonymous
  • anonymous
i did EXACTLY what you just did with the -17/7 and got those numbers.
anonymous
  • anonymous
I see that, but your answer for t=-17/7 showed a smaller numerator than denominator and it shouldd be the other way around.
anonymous
  • anonymous
then show me.
anonymous
  • anonymous
whats the y value for -17/7
anonymous
  • anonymous
How did you do that? Each numerator is larger than each denominator and they both have total of 7th powers.
anonymous
  • anonymous
ok find the answer.
anonymous
  • anonymous
i need to see if it's correct
anonymous
  • anonymous
i cant keep wasting 2-3 hours on one problem.
anonymous
  • anonymous
@calcmat, are you stuck?
anonymous
  • anonymous
\[\left(- \frac{ 24 }{ 7 } \right)^{4}\times \left( \frac{ 18 }{ 7} \right)\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
now 18/7 needs to be raised to the 3rd power.
anonymous
  • anonymous
ACCORDING TO THE ORIGINAL PROBLEM
anonymous
  • anonymous
Should gha have a 3 on the second expression for the power.
anonymous
  • anonymous
and i just proved my point. you are taking em to the wrong direction AGAIN.
anonymous
  • anonymous
Correct.
anonymous
  • anonymous
What are you talking about.
anonymous
  • anonymous
I ASKED YOU ONE SIMPLE QUESTION. CAN YOU FIND THE ANSWER?
anonymous
  • anonymous
I WANT TO SEE IF WHAT YOU'RE DOING IS ACTUALLY CORRECT
anonymous
  • anonymous
IVE SPENT 2-3 HOURS WITH YOU IN THIS PROBLEM.
anonymous
  • anonymous
Hey, don't get mad at me, I am the one spending MY TIME helping YOU.
anonymous
  • anonymous
YOURE STILL NOT ANSWERING MY QUESTION.
anonymous
  • anonymous
You can answer the question by taking (-24/7)^4(18/7)^3
anonymous
  • anonymous
I DID. STILL A HUGE NUMBER
anonymous
  • anonymous
I got 2349.5 rounded to 1 decimal place.
anonymous
  • anonymous
WRONG BUDDY.
dan815
  • dan815
(-1+t)^3 (5+t)^2 (17+7 t) your 0's = 1,-5,-17/7
anonymous
  • anonymous
nvm. i''ll solve it
anonymous
  • anonymous
thanks everyone.
anonymous
  • anonymous
Those are the ones I got @dan815
anonymous
  • anonymous
that part is correct.
anonymous
  • anonymous
and @calmat01 don't round. thats wrong.
anonymous
  • anonymous
keep the numbers just the way they are. p.s found the answer. thanks.
anonymous
  • anonymous
Ok, so my answer wasn't completely off?
dan815
  • dan815
mathcalculus can you prove the divergence theorem for me?
anonymous
  • anonymous
So you are telling me that you should have left the answer as 1934917632/823543? That's insane @mathcalculus
dan815
  • dan815
what are all these big numbers?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
lol insane.
anonymous
  • anonymous
your explanation was confusing. but the numbers are correct.
anonymous
  • anonymous
Gee, I spent all that time being patient with you, and all I get is my explanation was confusing?
anonymous
  • anonymous
lol truth to be told...truth be told.
anonymous
  • anonymous
I see. Well, if truth be told, the reason it was confusing was because you forgot how to factor. But I am not bitter. Take care and good luck with the rest of your assignment.
anonymous
  • anonymous
I did factor. I just didn't put it up because I was trying to understand you. And also, you were throwing out numbers without reason. & I'm glad you're not so bitter to take my words sensibly:) Thank you.

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