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

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ignore the 
 11 months ago

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

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

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

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

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

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