Find the inflexion points of the function x^x^x?

Question

Check this Y!A question:

http://answers.yahoo.com/question/index;_ylt=AhK9dd963eo7M_r0Jlzq6ML4xQt.?qid=20070820153043AAN8WHT

For positive x, how many inflexion points does this function have, and where? An inflexion point is when a function undergoes a change of curvature from positive to negative or vice-versa. Approximate answers accepted.

The function x^(x^x) is not the same as x^3.

Also, the function x^x^x is really supposed to be x^(x^x), not (x^x)^x.

Good work, Dr D and falzoon. Now, the first guy that figures out the OTHER inflexion point gets the 10 points.

Answer 1

From a graph, there seems to be only one obvious inflexion
point. I've estimated (with a lot of work), that the x value of
this point lies between 0.667657276 and 0.667657277,
and the y value is approx. 0.7050591553792515.
I hope to edit soon to let you know how I found that out.

EDIT:
First I took the derivative of y = x^(x^x) and got :
dy/dx = x^(x^x + x) {1/x + ln(x) [1 + ln(x) ] }

I then baulked rather horrendously at taking the second
derivative, so tried a different way.

From a graph, it is clear that the inflexion point has an
x value somewhere between 0.5 and 0.9. Up to the
inflexion point, the slope (dy/dx) is decreasing, and
afterwards, it is increasing. So I ran a little program to
find out at what x value was the minimum slope, by
inspection. On finding it, I then decreased the x range
and changed the step size and ran the program again,
and again, and again ... until I ran out of decimals in
double precision for the value of dy/dx. The program is
as follows (it could be programmed to do the whole lot
in one go, but that seemed like a lot of extra time) :
(Step size was chosen because only about 20 results
are able to be seen on the screen at the same time)

CLS : DEFDBL A-Z
FOR x = .5 TO .9 STEP .02
slope = x ^ (x ^ x + x) * (1 / x + LOG(x) + (LOG(x)) ^ 2)
PRINT x, slope
NEXT x

The next time around the FOR statement would be :
FOR x = .64 TO .68 STEP .002
etc. etc. for about 8 or 9 times.

EDIT 2:

Sorry, I made a mistake with my y value before.
What I gave you instead was the slope. ... !@#$%^& ...

To set the record straight, I give you the first inflexion
point as (0.66765728, 0.73456520).

ANOTHER inflexion point appears to be :
(0.639872, 0.714958).

EDIT 3:

Forgive me, I didn't mean to write that as the next point.
Missed a decimal and calculated the wrong y. (AGAIN!)
Those values are wrong.
The other inflexion point should be :
(0.06398720, 0.09969503)

Answer 2

Since you're linking to that question,
let y = (x^x)^x
lny = x^2 * lnx

Differentiating
y' / y = 2x*lnx + x

Again
(y*y" - (y')^2) / y^2 = 3 + 2lnx

At the point of inflexion, y" = 0
So -(y'/y)^2 = 3 + 2lnx
-[2x*lnx + x]^2 = 3+2lnx

Let f(x) = 3 + 2lnx + x^2 *[2lnx + 1]^2 = 0

That can be solved using Newton's method
x = 0.2022
y = 0.9367

*EDIT*
OK since you were l inking to that question I thought you meant the same def'n.
y = x^(x^x)
lny = x^x * lnx
Note d/dx (x^x) = x^x * (1 + lnx)
So y'/y = x^x * lnx * (1+lnx) + x^(x-1)

[y*y" - (y')^2]/y^2 =
x^x*lnx*(1+lnx)^2 + x^(x-1)*(2+3lnx - 1/x)

Again set y" = 0, LHS = -(y'/y)^2
Rearrange to get f(x) =
x^x*lnx*(1+lnx)^2 + x^(x-1)*(2+3lnx - 1/x)
+ [x^x * lnx * (1+lnx) + x^(x-1)]^2

Since we're solving numerically (Newton's method), there is no need to formally differentiate f(x). We can approximate
f'(x) = [f(x+0.001) - f(x)] / 0.001

Eventually
x = 0.66766
y = 0.73457

*EDIT*
For the other one, it works if you use x = 0.01 as a starting value for the Newton's method. And I"m getting
x = 0.06399
y = 0.09970

BUT Falzoon got it first.

Answer 3

Points of inflection are found by differentiating a function twice and setting it equal to zero.

Y = x^3 Take the derivative with respect to x (dy/dx)
Y' = 3x^2 Again, do the same/
Y"= 6x Set it equal to 0 and solve for x.
0 = 6x You can see this has one solution at x = 0.

There exists only one inflection point for this equation at x=0.

Answer 4

a ingredient on a mushy curve at which the curve differences its concavity from concave upward to concave downward or vice versa is termed a ingredient of inflection. factors of inflection ensue for values of x the place the 2nd spinoff of the function is equivalent 0. At a ingredient of inflection, the tangent to the curve crosses the curve, The tangent to a curve continually lies opposite the concave area of the curve. The W curve differences its concavity from concave as much as concave down and then back to concave up. So 2 factors of inflection. And 2 places wher f "(x) = 0.