extreme points of the function

[racecarace]

Find all extreme points of the function y=x^x^x, when x > 0

Please help?

 

[galactus]

Here is the graph?

If you want to differentiate, you can try logarithmic differentiation.

Do you know how to apply that?.

 

[racecarace]

I tried, haha.
I know how to do it once, for just x^x

but I'm getting confused with this one because the exponent is x^x again

If you take the ln of both sides
lny = lnx^(x^x)
Then bring the exponent down
lny = (x^x)lnx
Then take the derivative?
1/y y' = ?
How would you do that? Do you have to do logarithmic differentiation a second time?

 

[galactus]

I get:

$\displaystyle y=x^{(x^{x})}$

$\displaystyle ln(y)=ln(x^{(x^{x})})$

$\displaystyle ln(y)=x^{x}ln(x)$

Differentiate:

$\displaystyle \frac{y'}{y}=x^{x-1}\left(x(ln(x))^{2}+xln(x)\right)$

But, $\displaystyle y=x^{(x^{x})}$

Therefore, solving for y', we get:

$\displaystyle \boxed{y'=x^{x^{x}+x-1}\left(x(ln(x))^{2}+xln(x)+1\right)}$

This will be difficult to set to 0 and solve for x. Maybe try Newton's method. Or better yet, technology.

By looking at the graph, it doesn't appear to have many extrema. I tried Newton with various values and it didn't want to converge well for any of them.

 

[DR._Glockman]

Since x > 0 (given), the domain of the function is (0,?). Looking at the above graph of the function, it seems that no extrema exist, -- extrema being the minimum and/or maximum of a function on an interval.

f ' (x) = x^(x^2+x-1)[1+xln(x)+x[ln(x)]^2}. Set equal to zero. Since x raise to any power is always greater than zero, we only have 1+xln(x)+x[ln(x)]^2 =0. 1+xln(x)+x[ln(x)]^2 is always >0, hence, no extrema exist.

 

[galactus]

f ' (x) = x^(x^2+x-1)[1+xln(x)+x[ln(x)]^2}. Set equal to zero. Since x raise to any power is always greater than zero, we only have 1+xln(x)+x[ln(x)]^2 =0. 1+xln(x)+x[ln(x)]^2 is always >0, hence, no extrema exist.

Good observation.

I was certainly maker it tougher than need be talking about Newton and all that...duh.

 

[racecarace]

we only have 1+xln(x)+x[ln(x)]^2 =0. 1+xln(x)+x[ln(x)]^2 is always >0, hence, no extrema exist.

I see that it's always greater than zero, but why does that mean no extrema exist?

 

[racecarace]

Ohhhhh
Okay, just kidding I understand now, I wasn't thinking.
if its always greater than zero, it can never equal zero, so no critical points, no extrema...

But could you show me the steps on how you get the derivative?

f ' (x) = x^(x^2+x-1)[1+xln(x)+x[ln(x)]^2}.

 

[galactus]

But could you show me the steps on how you get the derivative?

f ' (x) = x^(x^2+x-1)[1+xln(x)+x[ln(x)]^2}.

I thought I showed that in my first post. What part do you not understand?.

To find the derivative of $\displaystyle x^{x}$ you can proceed as follows.

$\displaystyle y=x^{x}$

$\displaystyle ln(y)=ln(x^{x})=xln(x)$

$\displaystyle \frac{y'}{y}=ln(x)+x\cdot\frac{1}{x}=ln(x)+1$

$\displaystyle y'=y(ln(x)+1)=x^{x}(ln(x)+1)$

Now, use that with the product rule to find the derivative of $\displaystyle x^{x}ln(x)$

$\displaystyle x^{x}\cdot\frac{1}{x}+ln(x)(x^{x}(ln(x)+1))=x^{x-1}+ln(x)(x^{x}(ln(x)+1))=x^{x-1}\left(x(ln(x))^{2}+xln(x)+1\right)$

Now, when we multiply that by $\displaystyle x^{x^{x}}$ from the original problem above, we get the result we showed.