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#1 2006-01-06 00:49:42

3d_guru
Member
Registered: 2006-01-02
Posts: 5

derivative..

again..i've uploaded the picture because it's easier to write it in Mathematica..

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#2 2006-01-06 01:21:10

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: derivative..

If you use Mathematica it's easy.


IPBLE:  Increasing Performance By Lowering Expectations.

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#3 2006-01-06 01:25:50

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: derivative..

Put the following code:

Print["The Function:"]
f[x_]:=((x^2)Log[x])/ \[ExponentialE]
f[x]
Print["The Derivate:"]
f'[x]
Print["The roots:"]
Solve[f'[x] == 0, x]

IPBLE:  Increasing Performance By Lowering Expectations.

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#4 2006-01-06 01:29:52

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: derivative..

And you can include Plots:

Plot[f[x], {x, 0, 10}, AxesLabel -> {x, f[x]}]
Plot[f[x], {x, 0, 1}, AxesLabel -> {x, f[x]}]
Plot[f'[x], {x, 0, 10}, AxesLabel -> {x, f'[x]}]
Plot[f'[x], {x, 0, 1}, AxesLabel -> {x, f'[x]}]

IPBLE:  Increasing Performance By Lowering Expectations.

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#5 2006-01-06 01:39:34

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: derivative..

Here is the program:
(demonstrates the power of Mathematica once again)

Last edited by krassi_holmz (2006-01-06 01:41:21)


IPBLE:  Increasing Performance By Lowering Expectations.

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#6 2006-01-06 01:59:01

3d_guru
Member
Registered: 2006-01-02
Posts: 5

Re: derivative..

hey thanx krassi_holmz..didn't know Mathematica was so powerfull.
oh man and the code is so simple, that's great!

thanx again..i appreciate it

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#7 2006-01-06 02:14:38

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: derivative..

Haven't you read Mathematica help?

It's hundred times powerful than this. It's just a simple example.

Last edited by krassi_holmz (2006-01-06 02:15:32)


IPBLE:  Increasing Performance By Lowering Expectations.

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#8 2006-01-08 09:20:49

God
Member
Registered: 2005-08-25
Posts: 59

Re: derivative..

Derivative =0

x^2 = 0, x = 0

or

2xlnxe - lnx = 0
lnx(2xe-1) = 0

lnx = 0 or 2ex - 1 - 0
x = 1 or x = 1/(2e)

So, horizontal tangents occur at
x = 1
x = 1/(2e)
x = 0

Double check that the second derivative is never equal to 0 for each of those points... when it does, it isn't an extrema

Last edited by God (2006-01-08 09:21:56)

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