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Christos
 3 years 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
 3 years 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

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

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

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

Christos
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0(it's (6 ± sqrt(0)) / 2 )

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

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

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

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

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

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

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

reemii
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Why mistake what do you multiply first with what

Christos
 3 years 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

reemii
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yea I see now are my next moves correctly done??

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

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

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

reemii
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Bro I mean concave up down :D I moved a step

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

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

reemii
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0only for f'(x) and f''(x) ?

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

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

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

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

reemii
 3 years 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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