Question

I am going to do a concavity problem and abs max and min. I will just need a verification and corrections if anything.
(please, patience....)

Answer 1

ok

Answer 2

Given that: \(\large\color{black}{ f(x)=x^4-4x^3 }\)
So,

\(\large\color{black}{ f'(x)=4x^3-12x^2=4x^2(x-3) }\)
\(\large\color{black}{ f''(x)=12x^2-24x=12x(x-2) }\)

Critical numbers,
\(\large\color{black}{ 0=4x^2(x-3) }\)
\(\large\color{black}{ x=3,0 }\)

Answer 3

\(\large\color{black}{ f''(3)=0 }\) and \(\large\color{black}{ f'(3)>0 }\)
So, \(\large\color{black}{ f(3) }\) is the local minimum.

Answer 4

f''(x)=0, when x=0 or 2.

{-inf,0) concave up
(0,2) concave down
(2,0) concave up

Answer 5

the last interval should be (2,inf) and it is concave up

Answer 6

(0,0) is an inflection point, since the concavity changes there. (From up to down)

Answer 7

This is just a discussion question.

Answer 8

2,-16 is another inflection point, and the concavity changes from down to up.

Answer 9

ohh, ithe f(x) is on [0,4]

My bad sorry

Answer 10

I never post questions correctly-:(

Answer 11

Take your time; just let us know when you're ready for feedback. :-)

Answer 12

and 4 is also a critical number --- closed interval .

ready.

Answer 13

i AM DONE

Answer 14

the domain here does really matter?

Answer 15

are you looking for max, min, concavity what?

Answer 16

doesn't i mean

Answer 17

don't think the interval matters.

Answer 18

i do

Answer 19

I mean the [0,4] doesn't matter.

Answer 20

ii think it does

Answer 21

if you are looking for a max, you will not find one on \((-\infty,\infty)\) but you will certainly find one on \([0,4]\)

Answer 22

for the absolute mAXIMUM it does.

Answer 23

it would be when x=4

Answer 24

you are right, if you mean not for the minimum

Answer 25

the minimum is at x=3

Answer 26

But have I left anything out?

Answer 27

\[x^3(x-4)\] has a max of \(0\) at \(0\) and \(4\) on the closed interval

Answer 28

max of 0?

Answer 29

sure

Answer 30

\[f(0)=f(4)=0\]

Answer 31

Ohh, the maximum is 0 and 4

Answer 32

I see

Answer 33

oh no!

Answer 34

the maximum is the output, not the input

Answer 35

A couple things:

I think you meant to say that \(f^{\prime}(3) = 0\) and that \(f^{\prime\prime}(3) > 0\). Your conclusion that it's a minimum is correct (via the second derivative test).

We have no other critical points in the interval [0,4] since we don't allow critical points to be the endpoints of an interval. Now to determine absolute max and min, evaluate f(x) at the endpoints of the interval and at the sole critical point x=3:
\[\begin{array}{c|c} x & f(x) \\\hline 0 & 0 \\ 3 & -27\\ 4 & 0 \end{array}\]
Hence, the absolute max is 0 at x=0,4 and the absolute min is -27 at x=3.

Answer 36

I mean the max is at x=0 and at x=4

Answer 37

yes

Answer 38

yes, Christopher:)

Answer 39

tnx everyone:)
I have studied this second deriv. test and concavity the last lecture, but hope I'll be able to decently do these at some point.

Answer 40

Question closed.

Answer 41

there is a nice table that gathers all the info from f' and f"
critical point, points of inflaction...
much better to read!
like this
http://commons.wikimedia.org/wiki/File:Tableau_variation_polynome_1_3_1.png

Answer 42

My teacher prefers/requires (even if just prefers, I don't want to have a bad impression on him) it like a list of information

Critical numbers:
a, b, c, whatever they are

Concave up and down, in interval notations in order of the domain, if you know what I mean.
Like,
[2,3) concave up
(3,6) concave down
(6,7) concave up

like this in order of the x values.

see what I mean?
This si how I am supposed to enter my information on tests, h/w and discussions.

Answer 43

but ty though.

Answer 44

i see... well it depends on a person after all!

Answer 45

thats cool