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Topic review (newest first)

God
2006-01-09 08:20:49

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

krassi_holmz
2006-01-07 01:14:38

Haven't you read Mathematica help?

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

3d_guru
2006-01-07 00:59:01

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

krassi_holmz
2006-01-07 00:39:34

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

krassi_holmz
2006-01-07 00:29:52

And you can include Plots:

Code:

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]}]
krassi_holmz
2006-01-07 00:25:50

Put the following code:

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]
krassi_holmz
2006-01-07 00:21:10

If you use Mathematica it's easy.

3d_guru
2006-01-06 23:49:42

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

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