Daily Graph 10-04-09

Sorry i totally forgot that in {x} there is no negative part

{x}{x} ≤1 if x>0

For 0<x<1
y = xx

if y(xx)>x

log y =x log x
dy/ydx= 1 + logx
0 = y + y logx

0 =1 + logx
log x = -1

x = 1/2.72
x = 0.367 it it the minimum
y = 0.69 minimum

y=x^x in the interval [0,1)
y>0

let y=lim(x->0){x}^{x}
lny = lim(x->0){x}ln{x}
lny = lim(x->0+)lnx/(1/x) using LH rule
lny = lim(x->0+)(1/x)/(-1/x²)
lny = 0
y = 1

f(0+) = 1
f(1-) = 1

f'(x) = x^x + x^xlnx = x^x(1+lnx)
f''(x) = x^x(1+lnx) + (1+lnx)²x^x + x^x-1
= x^x-1[(1+lnx)(2+lnx) + 1]
= x^x-1[3+3lnx+(lnx)²] > 0

f'(x) >0 when 1+lnx>0 i.e. lnx > -1 i.e 1>x > 1/e
f'(x) <0 when x<1/e

Try dis 1

the limit would be 1?

because in 0-1

{x}=x

so it is equivalent to Lt_x→0x^x

sir

AM I WRONG IN SAYING THAT WE WON'T HAVE A VALUE OF THE FUNCTION AT THE INTEGERS
OR
THE VALUE EXISTS?????????/

I forgot the minus sign therefore reposted the graph

Bhaiyya pls check if its rite or wrong?