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mathcalculus
Group Title
HELP please. Locate all critical points :
(The answers are to be points.
 one year ago
 one year ago
mathcalculus Group Title
HELP please. Locate all critical points : (The answers are to be points.
 one year ago
 one year ago

This Question is Closed

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
Have you considered the 1st and 2nd derivatives?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay so i did this: s'(T)= 4(t3)^3 (t+2)^5+(t3)^4 5(t+2)^4(1)
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
Very good, excepting the upper case "T" on the left hand side. It is generally considered bad form to change variables in the middle of the equation. Okay, what does that 1st derivative tell us?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
wait: it is 4(t3)^3(1)(t2)^5+(t3)^4 5(t2)(1)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i am going to use product rule and chain rule.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
It is not necessary to carry the (1). It is understood or unnecessary. Okay, you have managed a correct 1st derivative twice. Now, what does that tell us?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
then... 4(t3)^3 (t2)^5 +5(t3)^4 (t2)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes but i still write it to be sure.. :/
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
To Luis_Rivera's "point", I think the intent is to report (a,b) rather than x = a.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
now where im stuck is on the factoring part...
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
well we know we have to find the points.... in order to do that we must find x, so both are right.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@tkhunny now how do i move on after this?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
That's fine. I have no objection to carrying the (1) along if it helps you understand what you are doing. Time to break out your algebra skills. 4(t3)^3 (t+2)^5 +5(t3)^4 (t+2)^4 Common (t3)^3 (t3)^3[ 4(t+2)^5 +5(t3)(t+2)^4] Common (t+2)^4 s'(t) = (t3)^3 (t+2)^4 [4(t+2) +5(t3)] Okay, now what? Note: I'm not going to do that for you on the 2nd Derivative.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
1) There is no (t2), it is (t+2). Please be more careful. 2) The common factor was (t+2)^4. This leaves one (t+2) in one term, since it started with (t+2)^5
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
Right now, we're just playing with symbols. It is important to be able to do that, but it doesn't help much if we don't know what to do with all the nice little symbols when we are done playing with them. I ask again. What does the 1st derivative tell us?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
anyone like to help? let me hagning up thereee;/
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.0
@mathcalculus when you work with a master, you would have to be attentive of what the question is about. If you don't understand the question, ask and ask again what he want to know from you. You are lucky to work with @tkhunny for the past halfan hour. If I have an advice at this point, make sure you understand the definition of "critical point" before proceeding, so you know what you're looking for!
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i've ACTUALLY been waiting..
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i do know the definition for critical point, what i needed help was on the factoring form..
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
part*
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
You haven't answered my question, so I was beginning to wonder if you were still working on it.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i calculated wrong. i needed to see where i made my mistake
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@tkhunny yes i was working on it.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@tkhunny that determines the critical points. the derivative can;t = to zero.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
Fair enough. It helps if you tell me, so I don't give up on you. If s'(t) = 0, what does that tell us? (t3)^3 = 0 at t = 3 (t+2)^4 = 0 at t = 2 [4(t+2) +5(t3)] = 0 at t = 7/9 So, it appears s'(t) CAN be zero. What do we know when it is?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
min or max?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
Maybe. Not if it is superseded by an inflection point. We must keep track. No conclusions until all the evidence is in.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
gtg Up to pdpilot326 from here.
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
not necessarily y = x^3 has a derivative f'(0) = 0 but it's not a min nor max.
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
check this... http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot is it okay to move on from where i left off?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
s'(t) = (t3)^3 (t+2)^4 [4(t+2) +5(t3)]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot326 thank you, i think I am understanding this better but i hate being asked questions like i'm some sort of mathematician..
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
the problem i has was factoring it.. now to find the critical points i know i have to look for x by setting it to zero after i find the derivative.
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
simplify in the brackets to better see the zero... \[[4\left( t+2 \right)+5\left( t3 \right)]=4t+8+5t15=9t7\] now you have all factors so any of them can be zero... \[s'\left( t \right)=\left( t3 \right)^{3}\left( t+2 \right)^{4}\left( 9t7 \right)\] so s'(t) = 0 if t = ?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
t=3 t=2 t=7/9
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
from there plug into original function and find the y's to each x
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
right?
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
yes, you can also check the 2nd derivative or use a number line with your critical points and check to see which intervals are + or . this will tell you if you have a max, min or neither.
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
\[s′(t)=(t−3)^{3}(t+2)^{4}(9t−7)\] dw:1375838972632:dw so you should have a rel. max at 7/9 and a rel. min at 3
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
but for this questions they want the points...
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i used the critical numbers and plugged into the orginal equation and i find it wrong.. it' seems wrong.
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
i understand... just giving you a bit extra.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
(3,0),(2,0),(7/9,4033.09)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh well thankyou for that.
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
it's not 7/9, it's 7/9...
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
7.9**
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ignore the 
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
no, 7/9 = 0.7777777777 to infinite and beyond...
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
oh, well then that's all good. why does it seem wrong?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
don't i use this critical number and plug it in?
 one year ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
yes, you want s(7/9)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh okay got cha!
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.1
So, we're not going to pursue the points of inflection?
 one year ago
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