# Exploring Slopes in f(x)=x^x^x and f(x)=x^x^x^x

• Rishav
In summary, the conversation discusses the function f(x)=x^x^x and its derivatives at various points. The slopes are found to be decreasing between 0-1 and increasing rapidly between 0.7-1. The conversation also briefly mentions another function f(x)=x^x^x^x and the process of differentiating it. The reason for the decreasing slopes is attributed to the nature of the function x^x being decreasing in that range.
Rishav
Hey people,
Consider the function:
f(x)=x^x^x
At 0,f(x)=0 and d/dx=1
At 1,f(x)=1 and d/dx=1
At 2,f(x)=16 and d/dx=107.11
At 3,f(x)=7.625597485000*10^12 and d/dx=5.43324993100000*10^14

Why this? Check the slopes...

And: Slopes between 0-1
At 0.1, f(x)=0.16057 d/dx= 1.658
At 0.2, f(x)=0.31146 d/dx=1.350
At 0.3, f(x)=0.43215 d/dx=1.070
At 0.4, f(x)=0.52987 d/dx=0.890
At 0.5, f(x)=0.61255 d/dx=0.777
At 0.6, f(x)=0.68662 d/dx=0.716
At 0.7, f(x)=0.75740 d/dx=0.708
At 0.8, f(x)=0.82972 d/dx=0.747
At 0.9, f(x)=0.90862 d/dx=0.840
At 1.0, f(x)=1.00000 d/dx=1.000
The slopes between 0.2-0.7 are decreasing...why's that? Can anyone explain?

And one more function f(x)=x^x^x^x
At 0, d/dx=-infinity
At 0.1, d/dx=-1.528330000
At 0.2, d/dx=-0.37292
At 0.3 d/dx~0
At 0.4 d/dx=0.31333
At 0.5 d/dx=0.45030
At 0.6 d/dx=0.54327
At 0.7 d/dx=0.63323
At 0.8 d/dx=0.72329
At 0.9 d/dx=0.83695
At 1.0 d/dx=1.000

So can anyone say why does the slope decrease and then increase so rapidly? Hope my questions are not falling on deaf ears!

Well, you might try to differentiate the animal and verify your observations!

arildno said:
Well, you might try to differentiate the animal and verify your observations!
Good one!

Is this ((x^x)^x)^x or (what I'm assuming) x^(x^(x^x))?

EDIT: Or I can just be smart and plug in one of the values you've so kindly given us. Never mind

It isn't that difficult to differentiate the creature.
Let's tackle the first one:
$$f(x)=x^{g(x)}=e^{\ln(x)g(x)}, g(x)=x^{x}=e^{x\ln(x)}$$
Thus,
$$\frac{df}{dx}=f(x)*(\frac{g(x)}{x}+\ln(x)g'(x))=f(x)*(\frac{g(x)}{x}+\ln(x)g(x)(\ln(x)+1))$$

This might then be written out more fully, if that's important to you.

Just plot the function, alongside x^x. The reason the slopes decrease is beacaues x^x is a decreasing function for a short time in and around that range.

## What is the difference between f(x)=x^x^x and f(x)=x^x^x^x?

The main difference between these two functions is the number of nested exponentials. In f(x)=x^x^x, there are three nested exponentials, while in f(x)=x^x^x^x, there are four nested exponentials.

## What is the domain and range of these functions?

The domain of both functions is all real numbers greater than 0. However, the range of f(x)=x^x^x is also all real numbers greater than 0, while the range of f(x)=x^x^x^x is all real numbers greater than or equal to 1.

## How do these functions behave as x approaches infinity?

As x approaches infinity, both functions approach infinity. However, f(x)=x^x^x grows at a faster rate than f(x)=x^x^x^x, so it will have a steeper slope.

## Do these functions have any asymptotes?

Neither of these functions have any vertical or horizontal asymptotes. However, they do have a slant asymptote at y=x, where the slope of the function approaches 1 as x approaches infinity.

## Can these functions be graphed on a standard coordinate plane?

Yes, both functions can be graphed on a standard coordinate plane. However, as x approaches infinity, their graphs will quickly become too large to be displayed on a standard scale.

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