Can you check if this is correct?

J

Joey29

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Dec 31, 2005
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Hey, can anyone check if this is done properly? I'm trying to find out where the function is concave up or down. Can you check if this is correct (method & math)?


k(x) = -2x^3 (x-4)
k'(x)= -8x^3+24x^2

K''(x) = -24x^2+ 48x = 0
-24x^2 +48x
-24x(x+2) = 0
x= 0
x= -2

^-----inflation points, right?

----neg---neg---pos
-------- -2-----0-----
--CD------CD---CU

k''(-3)= -360..
k"(-1)= -72.00..
k"(1)= 23.99...

so it would be concave down from - infinite to -2, concave down from -2, 0 and concave up from 0 to infinite? Would that be correct? I am not certain if the procedure is legal. I appreciate any help. Can anyone sketch a graph of this fn, I sketched one on paper but I do not know how to sketch one here. It would be helpful as well to see if my graph is correct (Ignore y-intercepts).

Joe
 
G'day, Joe.

Joey29 said:
Hey, can anyone check if this is done properly? I'm trying to find out where the function is concave up or down. Can you check if this is correct (method & math)?


k(x) = -2x^3 (x-4)
k'(x)= -8x^3+24x^2

K''(x) = -24x^2+ 48x = 0
-24x^2 +48x
-24x(x+2) = 0 <- Minor error. That should be -24x(x - 2) = 0
x= 0
x= -2 So we actually have x=2

^-----inflation points, right?
We always have to check that there is indeed a change in sign of the second derivative. For example, y=x^4 has a second derivative of zero at x=0, but no change in the sign of the second derivative / concavity occurs.

The question wants us to find the intervals of concavity, so we'll find out if those are inflexion points anyway.

----neg---neg---pos
-------- -2-----0-----
--CD------CD---CU

k''(-3)= -360..
k"(-1)= -72.00..
k"(1)= 23.99...

It may be useful to format it something like:
Code: 
                                     suspect inflection
                                       points     
                                     |        \ 
                                     |         \
                                    \|/        _\|
*---------------+-----+-----+-----+-----+-----+-----+-----+-----*
|           x   |-100 | -2  | -1  |  0  |  1  |  2  |  3  | 100 |
+---------------+-----+-----+-----+-----+-----+-----+-----+-----+
|Sign of k''(x) |   - |  -  |  -  |  0  |  +  |  0  |  -  |  -  |
*---------------+-----+-----+-----+-----+-----+-----+-----+-----*


The -100 and 100 are there if we're a little unsure of ourselves.

From the chart, the function is concave down (ie. k''(x) is negative)
in \(\displaystyle (-\infty, 0)\) and \(\displaystyle (0, +\infty)\).
And concave up (ie. k''(x) is positive) in \(\displaystyle (0, 2)\).
Click to expand...
 
Thank You!

Thanks Unco, that was a hard mistake to find yet you found it! It led to the outcome being wrong! I appreciate your help!

Joe
 
Hello, Joe!

Unco already caught your tiny error.
\(\displaystyle \;\;\;\)I'll just edit your statements.

\(\displaystyle k(x)\;=\;-2x^4\,+\,8x^3\)

\(\displaystyle k'(x)\;=\;-8x^3\,+\,24x^2\)

\(\displaystyle k''(x)\;=\;-24x^2\,+\,48x\;=\;0\)

\(\displaystyle \;\;\;-24x(x\,-\,2)\:=\:0\;\;\Rightarrow\;\;x\,=\,0,\,2\)


\(\displaystyle k''(-1)\,=\,-72,\;\;\;k(1)''\,=\,+24,\;\;\;k''(3)\,=\,-24\)

\(\displaystyle \;\;- - - 0 - - - - 2 - - -\)
. . .neg\(\displaystyle \;\;\;\;\)pos\(\displaystyle \;\;\;\;\)neg
. . . .\(\displaystyle \cap\;\;\;\;\;\cup\;\;\;\;\;\:\cap\)


Concave down on \(\displaystyle (-\infty,\,0)\) and \(\displaystyle (2,\,\infty)\)
Concave up on \(\displaystyle (0,\,2)\)


From the first derivative, we find a critical value at \(\displaystyle x = 3\)
\(\displaystyle \;\;\;\)It turns out to be a maximum at \(\displaystyle (3,\,54)\)

And we have inflection points at \(\displaystyle (0,\,0)\) and \(\displaystyle (2,\,32)\).
Code: 
          |         ***
          |      *   :  *
          |     *    :    *
          |   * :    :     *
   -------*-----+----+------
      *   |     2    3      *
    *     |
   *      |
The graph probably looks like this . . .
 
More than I can ask for

Thanks Soroban! Thats more than I can ask for. You mastered the LaTex language. I can barely get a fraction over a fraction in the LaTex format. I actually worked on it for a while the other day. I even stuck paper tape to my monitor with LaTex commands like you. Thanks again! The graph was similar to the one I created on paper:)
 
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