Let's analyze the given function $f(x) = (x - 2)^{1/5}$ over the interval $[0, 3]$ to determine if the Lagrange Mean Value Theorem (LMVT) is applicable.

The Mean Value Theorem (MVT) requires two conditions to be satisfied for a function $f$ on an interval $[a, b]$:

1. Continuity: $f(x)$ must be continuous on the closed interval $[a, b]$.
2. Differentiability: $f(x)$ must be differentiable on the open interval $(a, b)$.

Step 1: Continuity on $[0, 3]$

The function $f(x) = (x - 2)^{1/5}$ is a real-valued function. To check its continuity, note that the function is continuous everywhere except at the point where $x - 2 = 0$, i.e., $x = 2$. The function is continuous over the entire real line since fractional exponents like $(x-2)^{1/5}$ are defined for all real numbers. Therefore, $f(x)$ is continuous on the entire interval $[0, 3]$.

Step 2: Differentiability on $(0, 3)$

Now, we need to check differentiability on the open interval $(0, 3)$.

• The derivative of $f(x)$ is: $f'(x) = \frac{1}{5} (x - 2)^{-4/5}$ This derivative is not defined at $x = 2$, because the exponent $-4/5$ causes a division by zero when $x = 2$. Therefore, $f(x)$ is not differentiable at $x = 2$, even though it is differentiable everywhere else in the interval.

Conclusion:

Since the function $f(x) = (x - 2)^{1/5}$ is not differentiable at $x = 2$, the Mean Value Theorem is not applicable on the interval $[0, 3]$.

Would you like more details on any of the steps?

Here are 5 related questions to explore further:

1. What is the Lagrange Mean Value Theorem and how is it applied?
2. How do you determine whether a function is continuous or differentiable?
3. Can MVT apply to functions with fractional exponents?
4. What happens when a function is not differentiable at a point in the interval?
5. How would MVT be applied if we restricted the interval to avoid the non-differentiable point?

Tip: When checking for the applicability of MVT, always ensure differentiability across the entire open interval, not just at the endpoints.