The interval of the root of equation X-1 = lgx

The interval of the root of equation X-1 = lgx

The equation is transformed into f (x) = x-1-lgx
Using dichotomy, when f (x1) * f (x2)

The interval of the zero point of function f (x) = lgx-1 / X is a (0,1) B (1,10) C (10100) d (100, + 00)

f(1)=-10
f(100)=2-1/100>0
So the zero point is at (1,10)
Choose B

F (2 / x + 1) = lgx, find f (x)

Let 2 / x + 1 = t, then x = 2 / (t-1)
Then f (T) = LG ((2 / (t-1)) = LG2 LG (t-1)
That is, f (x) = lg2-lg (x-1)

F (x) = f (1 / x) lgx ^ 2 + 1 to find f (x)
I found out that (2lgx + 1) / (4 (lgx) ^ 2 + 1) the teacher crossed and didn't know what was wrong

Let t = 1 / x, x = 1 / T, and substitute it into the original formula
f(1/t)=f(t)lg(t^(-2))+1=-2f(t)lgt+1
In this case, replace T with X
We get f (1 / x) = - 2F (x) * lgx + 1
Substituting in the original form, we get
f(x)=[-2f(x)*lgx+1]*2lgx+1=-4(lgx)^2*f(x)+2lgx+1
Well organized
f(x)=[2lgx+1]/[1+4(lgx)^2]