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Christos
Group Title
Graphing derivatives , and derivative's characteristics
http://screencast.com/t/GxfCh0DY
So basically this exercise is 100% to be in our final exam today at Calculus , as stated by our teacher so I am trying to master it
 one year ago
 one year ago
Christos Group Title
Graphing derivatives , and derivative's characteristics http://screencast.com/t/GxfCh0DY So basically this exercise is 100% to be in our final exam today at Calculus , as stated by our teacher so I am trying to master it
 one year ago
 one year ago

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reemii Group TitleBest ResponseYou've already chosen the best response.0
and how's it going until now?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
For example On (a) do I have to find roots for only f(x) or f'(x) and f''(x) ?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
it says find roots of f
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
the roots of f: the x's such that f(x)=0 they are (graphically) the x's where the graph crosses the xaxis
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
ooh first = 3/2
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
second,third = (6 + sqrt(72))/2
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Which brings me to another problem of mine on how to actually point this number on the graph
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
not sure. in the form \((x+3)^2\) you should not compute the square, actually they already made the factorization for you. you can just read the answer: second and third roots are equal: x=3.
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
(it's (6 ± sqrt(0)) / 2 )
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
ah yea that's true
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
so we have one root for 3/2 and one for 3
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
for b I just find f(0) of the function ?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
yes again ;)
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
So the intervals (inf,9) increase (8,0) decrease (0,inf) increase At least thats what I got
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
the derivative of the function is \(6x(x+3)\). the intervals are (inf,3), (3,0), (0,inf)
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
arent we using multiplication rule for the derivative
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
yes. whether the values you found are right or wrong, at least verify that the intervals don't leave gaps. (your intervals left a gap between 9 and 8)
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
or was it a typo?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
it wasnt a typo but I guess you are right on that one
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
directly using the mult. rule: \(((2x3)(x+3)^2)' = 2(x+3)^2 + (2x3)2(x+3) = 2(x+3) \times [(x+3)+(2x3)]\).
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
oops. last one is \[ 2(x+3)\times [(x+3) + (2x3)] = 2(x+3)[3x] = 6x(x+3)\]
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
hm let me redo it
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
2(x+3)^2 +4x^2 + 12x  4x 12 up until now all correct?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
i don't think so. do you apply the rule on \((2x3)(x^2+6x+9)\) ? what's your starting point?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Yes I apply the rule however i start with the product rule first before applying it
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
show the first steps plz
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
2(x+3)^2+2(x+3(2x3)
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
2(x^2+6x+9)+2(x+3)(2x3)
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
those are correct?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
yes, you made a mistake when computing the product \(2(x+3)(2x3)\). this is equal to \(4x^2+6x18\).
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Why mistake what do you multiply first with what
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
ok ok I got it now for concave up/down (inf,1) down (1,inf) up
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
infection points x=1
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
and I am stuck at (f)
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
your derivation is correctly done. then it's only multiplication: \[2(x^2+6x+9) + 2(x+3)(2x3)\\ \quad = 2x^2+12x+18 + 2(2x^23x+6x9) \\ \quad = 2x^2+12x+18 + 4x^2 + 6x  18\\ \quad = 6x^2 + 18x \\ \quad = 6x(x+3)\]> roots are 3 and 0. > (inf, 3), (3,0), (0,inf) are the intervals.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Yea I see now are my next moves correctly done??
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
you didn't obtain the correct expression for f' (you got the wrong roots > wrong intervals). the inflexion point is at 3/2.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
you mean for f''(x) I got it wrong?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
I damn you are right
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
\(f'\) is unfortunately wrong, so you could not get the right \(f''\).
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
(inf, 3/2) down (3/2, inf) up I get it now
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
ouch. no too fast. \(f'(x) = 6x(x+3)\) with roots 3 and 0. > (inf,3) up, (3,0) down, (0,+inf) up.
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
\(f''(x) = (6(x^2+3x))' = 6(2x+3)\) with root x=3/2. > inflexion point at 3/2.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Bro I mean concave up down :D I moved a step
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
I understood the previews one
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
I am just stuck at (f) could you help me out surpass it?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
oh ok. (about concave up down) relative extrema means you have to study what happens at each x such that f'(x)=0, at 3 and at 0 in our case.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
only for f'(x) and f''(x) ?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
the relative extrema (local/global min or max) are only possibly located at the roots of \(f'(x)\). Look at \(x=3\) and say if it's a min, max, or none of these (inflexion point). Do the same with \(x=0\). You can use \(f''\) to tell you about the shape of the curve at that particular point.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
is not a min nor a max at 3 0
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
you mean (3,0)? it is a local maximum. Reason: \(f''(3)<0\). (or just \(f\) is increasing before 3 and decreasing after 3.).
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
bro you told me to check at f'(x) so which one do I check? do I check both f'(x) and f''(x) ? Do I only use the roots from the first derivative?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
1) to answer the question you have to do some job at each of the points that solve \(f'(x)=0\). We know that these points are 3 and 0. 2.1) what happens at 3 ? 2.2) what happens at 0 ? To answer 2.1 and 2.2, you can use any means you want. For example, a quick way is to use \(f''\). \(f''(3)>0\) means it's a MIN. \(f''(3)<0\) means it's a MAX. \(f''(3)=0\) means it's an inflexion point.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
so I dont use 3/2 at all? I can only just use the roots of the first derivative for the second?
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
What we know is that 3/2 is an inflexion point. But it happens (often?) that it is not an extrema. Drawing: dw:1369665423749:dw
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
so you don't need to look at it at all in this question (f). You would have to look at it if it was solution of \(f'(x)=0\). But it is NOT.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Ok and the last and hardest part for me can you help me analise/make the graph?? Would it be possible??
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
\(x=3\) > \(f''(3) = 6(2*(3)+3) < 0\) > local max \(x=0\) > \(f''(0) = 6(0 + 3)>0\) > local min. Ok
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
1) Always start with : put points of \(f\) that are easy: the roots, the intercept. 2) draw the missing part using the info in \(f'\) and \(f''\).
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Kinda can you show me what you do to graph it? I have like 5 minutes before the exams that's the most I can get atm
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
first step: a few points dw:1369666161790:dw
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
then join using use \(f''\) at the extrema (3 and 0) dw:1369666278751:dw
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
join in a healthy way (without forgetting hte inflexion point) dw:1369666362664:dw
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
use that order: 1)fixed points, 2)concavity at extrema 3) join. (pay attention to inflexion point, but that is less important than the rest)
 one year ago
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