Question

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

Answer 1

and how's it going until now?

Answer 2

For example On (a) do I have to find roots for only f(x) or f'(x) and f''(x) ?

Answer 3

it says find roots of f

Answer 4

the roots of f:
the x's such that f(x)=0
they are (graphically) the x's where the graph crosses the x-axis

Answer 5

ooh first = 3/2

Answer 6

second,third = (-6 +- sqrt(72))/2

Answer 7

Which brings me to another problem of mine on how to actually point this number on the graph

Answer 8

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.

Answer 9

(it's (-6 ± sqrt(0)) / 2 )

Answer 10

ah yea that's true

Answer 11

so we have one root for 3/2 and one for -3

Answer 12

for b I just find f(0) of the function ?

Answer 13

yes

Answer 14

-27

Answer 15

yes again ;)

Answer 16

So the intervals (-inf,-9) increase
(-8,0) decrease
(0,inf) increase

At least thats what I got

Answer 17

the derivative of the function is \(6x(x+3)\). the intervals are (-inf,-3), (-3,0), (0,inf)

Answer 18

arent we using multiplication rule for the derivative

Answer 19

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)

Answer 20

or was it a typo?

Answer 21

it wasnt a typo but I guess you are right on that one

Answer 22

directly using the mult. rule:
\(((2x-3)(x+3)^2)' = 2(x+3)^2 + (2x-3)2(x+3) = 2(x+3) \times [(x+3)+(2x-3)]\).

Answer 23

oops. last one is
\[ 2(x+3)\times [(x+3) + (2x-3)] = 2(x+3)[3x] = 6x(x+3)\]

Answer 24

hm let me redo it

Answer 25

2(x+3)^2 +4x^2 + 12x - 4x -12 up until now all correct?

Answer 26

i don't think so.
do you apply the rule on \((2x-3)(x^2+6x+9)\) ? what's your starting point?

Answer 27

Yes I apply the rule however i start with the product rule first before applying it

Answer 28

show the first steps plz

Answer 29

2(x+3)^2+2(x+3(2x-3)

Answer 30

2(x^2+6x+9)+2(x+3)(2x-3)

Answer 31

those are correct?

Answer 32

yes, you made a mistake when computing the product \(2(x+3)(2x-3)\). this is equal to \(4x^2+6x-18\).

Answer 33

Why mistake what do you multiply first with what

Answer 34

ok ok I got it now for concave up/down (-inf,-1) down (-1,inf) up

Answer 35

infection points x=-1

Answer 36

and I am stuck at (f)

Answer 37

your derivation is correctly done.
then it's only multiplication:
\[2(x^2+6x+9) + 2(x+3)(2x-3)\\
\quad = 2x^2+12x+18 + 2(2x^2-3x+6x-9) \\
\quad = 2x^2+12x+18 + 4x^2 + 6x - 18\\
\quad = 6x^2 + 18x \\
\quad = 6x(x+3)\]-> roots are -3 and 0.
-> (-inf, -3), (-3,0), (0,inf) are the intervals.

Answer 38

Yea I see now are my next moves correctly done??

Answer 39

you didn't obtain the correct expression for f' (you got the wrong roots -> wrong intervals). the inflexion point is at -3/2.

Answer 40

you mean for f''(x) I got it wrong?

Answer 41

I damn you are right

Answer 42

\(f'\) is unfortunately wrong, so you could not get the right \(f''\).

Answer 43

(-inf, -3/2) down (-3/2, inf) up I get it now

Answer 44

right?

Answer 45

ouch. no too fast.

\(f'(x) = 6x(x+3)\) with roots -3 and 0. -> (-inf,-3) up, (-3,0) down, (0,+inf) up.

Answer 46

\(f''(x) = (6(x^2+3x))' = 6(2x+3)\) with root x=-3/2. -> inflexion point at -3/2.

Answer 47

Bro I mean concave up down :D I moved a step

Answer 48

I understood the previews one

Answer 49

I am just stuck at (f) could you help me out surpass it?

Answer 50

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.

Answer 51

only for f'(x) and f''(x) ?

Answer 52

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.

Answer 53

is not a min nor a max at -3 0

Answer 54

you mean (-3,0)?
it is a local maximum. Reason: \(f''(-3)<0\). (or just \(f\) is increasing before -3 and decreasing after -3.).

Answer 55

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?

Answer 56

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.

Answer 57

so I dont use -3/2 at all? I can only just use the roots of the first derivative for the second?

Answer 58

@reemii

Answer 59

What we know is that -3/2 is an inflexion point.
But it happens (often?) that it is not an extrema. Drawing: