Lagrange's Mean Value Theorem Why is theta between 0 and 1?

Lagrange's Mean Value Theorem Why is theta between 0 and 1?

Sometimes it's a little hard to answer
θ belongs to (0,1)
Let's talk about Lagrange's mean value theorem first
If a function is continuous on [a, b] and differentiable on (a, b), then there is at least one point e between (a, b), which has:
f(b)-f(a)=f'(e)(b-a) (*)
(*) this formula can also be written as:
f(b)-f(a)=f'(a+θ(b-a))*(b-a)
Look at this: does a + θ (B-A) belong to (a, b)
Yes. Because θ belongs to (0,1), so: a + θ (B-A) belongs to (a, b)
In fact, you don't have to worry about the finite increment formula. Khan, let's understand it this way, e is a to B
When θ = 0, e is a, and when θ = 1, e is B
A + θ (B-A) always changes between (a, b)
The more I feel, the more I confuse you

What is Lagrange's theorem?

If the function f (x) is differentiable on (a, b) and continuous on [a, b], then there must be a ξ ∈ [a, b] such that
f'(ξ)*(b-a)=f(b)-f(a)

Lagrange's Mean Value Theorem
It is proved that parabola always accords with mean value theorem in any interval
I don't understand. Where did K (x1 ^ 2-x2 ^ 2) / (x1-x2) come from????

Let the parabola be y = KX ^ 2 (k is a non-zero constant), a (x1, Y1), B (X2, Y2) be any two points on the parabola
y'=2kx=(y1-y2)/(x1-x2)=k(x1^2-x2^2)/(x1-x2)=k(x1+x2)
x=(x1+x2)/2
So the proposition holds
Supplement: a (x1, Y1), B (X2, Y2) are any two points on the parabola, so Y1 = kx1 ^ 2, y2 = kx2 ^ 2