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Christos
 one year ago
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
Christos
 one year ago
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

This Question is Closed

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

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

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

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

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

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0not 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.

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

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

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

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

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

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0yes. 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)

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

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

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

Christos
 one year ago
Best ResponseYou've already chosen the best response.02(x+3)^2 +4x^2 + 12x  4x 12 up until now all correct?

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

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

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

Christos
 one year ago
Best ResponseYou've already chosen the best response.02(x^2+6x+9)+2(x+3)(2x3)

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

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

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

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

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0your 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.

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

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

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

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

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

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

reemii
 one year ago
Best 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.

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

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

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0oh 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.

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0the 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.

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

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

Christos
 one year ago
Best ResponseYou've already chosen the best response.0bro 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?

reemii
 one year ago
Best ResponseYou've already chosen the best response.01) 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.

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

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0so 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.

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

reemii
 one year ago
Best 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

reemii
 one year ago
Best ResponseYou've already chosen the best response.01) 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 \(f''\).

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Kinda 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

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

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

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

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