The Lagrange Mean Value Theorem (LMVT), also known as the Mean Value Theorem (MVT), states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $c \in (a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

To apply LMVT, two conditions must be satisfied:

1. Continuity on the closed interval $[a, b]$.
2. Differentiability on the open interval $(a, b)$.

Now, let's analyze the function $F(x) = |x|$ on the interval $[0, 1]$:

1. Continuity:
   The absolute value function $F(x) = |x|$ is continuous on the interval $[0, 1]$ because the absolute value function is continuous everywhere.

2. Differentiability:
   The function $F(x) = |x|$ is differentiable on the open interval $(0, 1)$, but not differentiable at $x = 0$. At $x = 0$, there is a sharp corner in the graph of $F(x)$, so the derivative does not exist at that point.

Conclusion:

Since $F(x) = |x|$ is not differentiable at $x = 0$, we cannot apply LMVT on the interval $[0, 1]$ because one of the conditions (differentiability on the entire open interval) is not met.

Do you need more details or have any questions?

Here are 5 related questions you might find interesting:

1. What happens if we apply LMVT on an interval where the function is not differentiable?
2. Can the Mean Value Theorem be applied on non-differentiable functions at points?
3. How is LMVT different from Rolle’s Theorem?
4. Can LMVT be applied on the absolute value function on a different interval?
5. What are the implications of non-differentiability on the existence of a tangent line?

Tip: When checking if you can apply LMVT, always verify the differentiability of the function across the entire open interval, not just continuity.