HELP please. Locate all critical points :
(The answers are to be points.

- anonymous

HELP please. Locate all critical points :
(The answers are to be points.

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- anonymous

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- tkhunny

Have you considered the 1st and 2nd derivatives?

- anonymous

okay so i did this:
s'(T)= 4(t-3)^3 (t+2)^5+(t-3)^4 5(t+2)^4(1)

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## More answers

- tkhunny

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?

- anonymous

wait: it is 4(t-3)^3(1)(t-2)^5+(t-3)^4 5(t-2)(1)

- anonymous

i am going to use product rule and chain rule.

- tkhunny

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?

- anonymous

then... 4(t-3)^3 (t-2)^5 +5(t-3)^4 (t-2)

- anonymous

yes but i still write it to be sure.. :/

- tkhunny

To Luis_Rivera's "point", I think the intent is to report (a,b) rather than x = a.

- anonymous

now where im stuck is on the factoring part...

- anonymous

well we know we have to find the points.... in order to do that we must find x, so both are right.

- anonymous

@tkhunny now how do i move on after this?

- tkhunny

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(t-3)^3 (t+2)^5 +5(t-3)^4 (t+2)^4
Common (t-3)^3
(t-3)^3[ 4(t+2)^5 +5(t-3)(t+2)^4]
Common (t+2)^4
s'(t) = (t-3)^3 (t+2)^4 [4(t+2) +5(t-3)]
Okay, now what?
Note: I'm not going to do that for you on the 2nd Derivative.

- tkhunny

1) There is no (t-2), 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

- anonymous

ok

- tkhunny

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?

- anonymous

anyone like to help? let me hagning up thereee;/

- mathmate

@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 half-an 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!

- anonymous

i've ACTUALLY been waiting..

- anonymous

i do know the definition for critical point, what i needed help was on the factoring form..

- anonymous

part*

- tkhunny

You haven't answered my question, so I was beginning to wonder if you were still working on it.

- anonymous

i calculated wrong. i needed to see where i made my mistake

- anonymous

@tkhunny yes i was working on it.

- anonymous

@tkhunny that determines the critical points. the derivative can;t = to zero.

- tkhunny

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?
(t-3)^3 = 0 at t = 3
(t+2)^4 = 0 at t = -2
[4(t+2) +5(t-3)] = 0 at t = 7/9
So, it appears s'(t) CAN be zero. What do we know when it is?

- anonymous

min or max?

- tkhunny

Maybe. Not if it is superseded by an inflection point. We must keep track. No conclusions until all the evidence is in.

- tkhunny

gtg Up to pdpilot326 from here.

- anonymous

not necessarily
y = x^3 has a derivative f'(0) = 0 but it's not a min nor max.

- anonymous

check this...
http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx

- anonymous

@pgpilot is it okay to move on from where i left off?

- anonymous

yep

- anonymous

s'(t) = (t-3)^3 (t+2)^4 [4(t+2) +5(t-3)]

- anonymous

@pgpilot326 thank you, i think I am understanding this better but i hate being asked questions like i'm some sort of mathematician..

- anonymous

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.

- anonymous

simplify in the brackets to better see the zero...
\[[4\left( t+2 \right)+5\left( t-3 \right)]=4t+8+5t-15=9t-7\]
now you have all factors so any of them can be zero...
\[s'\left( t \right)=\left( t-3 \right)^{3}\left( t+2 \right)^{4}\left( 9t-7 \right)\]
so s'(t) = 0 if t = ?

- anonymous

t=3 t=-2 t=7/9

- anonymous

from there plug into original function and find the y's to each x

- anonymous

right?

- anonymous

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.

- anonymous

\[s′(t)=(t−3)^{3}(t+2)^{4}(9t−7)\]
|dw:1375838972632:dw|
so you should have a rel. max at 7/9 and a rel. min at 3

- anonymous

but for this questions they want the points...

- anonymous

i used the critical numbers and plugged into the orginal equation and i find it wrong.. it' seems wrong.

- anonymous

i understand... just giving you a bit extra.

- anonymous

(3,0),(-2,0),(-7/9,4033.09)

- anonymous

oh well thankyou for that.

- anonymous

it's not -7/9, it's 7/9...

- anonymous

7.9**

- anonymous

ignore the -

- anonymous

no, 7/9 = 0.7777777777 to infinite and beyond...

- anonymous

O_o

- anonymous

oh, well then that's all good. why does it seem wrong?

- anonymous

don't i use this critical number and plug it in?

- anonymous

yes, you want s(7/9)

- anonymous

oh okay got cha!

- tkhunny

So, we're not going to pursue the points of inflection?

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