A proof of Lagrange's mean value theorem Verification: when x > 0, there is 1 / (1 + x)

A proof of Lagrange's mean value theorem Verification: when x > 0, there is 1 / (1 + x)

According to Lagrange's mean value theorem, there is a point x0 in the interval [x, x + 1], so that
(lnx0)' = ln(x+1) - lnx
Namely
1/x0 = ln(1+1/x)
0

The problem of the domain of the function in the first year of senior high school,
If the domain of function y = f (2x-1) is [0,1], then the domain of function y = f (1-3x) is ()
The answer is (0,2 / 3]

Because the domain is [0,1)
So the value of 2x-1 is [- 1,1)
So the value of 1-3x [- 1,1)
So the domain of definition of function y = f (1-3x) is (0,2 / 3]

The domain of F (x + 1) is [- 2.3]
Finding the domain of F (2x-1)

X belongs to [- 2,3]
X + 1 belongs to [- 1,4]
2x-1 belongs to [- 1,4]
2X belongs to [0,5]
X belongs to [0,2.5]

1. Given that the domain of F (x) is [- 2,1], find the domain of F (3x-1)
2. Given that the domain of F (2x + 5) is [- 1,4], find the domain of F (x)
Just write a brief process. The answer should be correct. Don't answer if you're not sure,

The function f (3x-1) can be divided into
y=f(t)
t=3x-1
F (T) and f (x) are the same function
-2≤t≤1
-2≤3x-1≤1
-1/3≤x≤2/3
therefore
F (3x-1) is defined as: [- 1 / 3,2 / 3]
two
The function f (2x + 5) can be divided into two parts;
y=f(t)
t=2x+5
Because - 1 ≤ x ≤ 4
3≤2x+5≤13
Namely
3≤t≤13
That is to say, the definition field of function f (T) is [3,13]
Because f (T) and f (x) are the same function, so
The domain of F (x) is [3,13]