- #1
Rishav
- 1
- 0
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!
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!