Verify Lagrange's Mean Value Theorem for the function :
`f(x)={{:(2+x^(3)" if "xle1),(3x" if "xgt1):}` on [-1, 2].

Answer

To verify Lagrange's Mean Value Theorem (LMVT) for the given piecewise function:

\[ f(x) = \begin{cases} 2 + x^3 & \text{if } x \leq 1 \\ 3x & \text{if } x > 1 \end{cases} \]

on the interval \([-1, 2]\), we will follow these steps:

### Step 1: Check Continuity on \([-1, 2]\)

1. **Evaluate \(f(-1)\)**:
\[ f(-1) = 2 + (-1)^3 = 2 - 1 = 1 \]

2. **Evaluate \(f(2)\)**:
\[ f(2) = 3 \cdot 2 = 6 \]

3. **Evaluate \(f(1)\)**:
\[ f(1) = 2 + 1^3 = 2 + 1 = 3 \]

4. **Check the limit from the left at \(x = 1\)**:
\[ \lim_{h \to 0^-} f(1 + h) = f(1) = 3 \]

5. **Check the limit from the right at \(x = 1\)**:
\[ \lim_{h \to 0^+} f(1 + h) = f(1 + h) = 3(1 + h) \to 3 \text{ as } h \to 0 \]

Since \(f(1) = 3\) and both limits equal \(3\), \(f(x)\) is continuous at \(x = 1\). Therefore, \(f(x)\) is continuous on \([-1, 2]\).

### Step 2: Check Differentiability on \([-1, 2]\)

1. **Differentiate \(f(x)\)**:
- For \(x < 1\):
\[ f'(x) = 3x^2 \]
- For \(x > 1\):
\[ f'(x) = 3 \]

2. **Evaluate \(f'(1)\)**:
- From the left:
\[ f'(1) = 3(1)^2 = 3 \]
- From the right:
\[ f'(1) = 3 \]

Since \(f'(1)\) exists and is equal from both sides, \(f(x)\) is differentiable on \([-1, 2]\).

### Step 3: Apply Lagrange's Mean Value Theorem

According to LMVT, there exists at least one \(c \in (-1, 2)\) such that:

\[ f'(c) = \frac{f(2) - f(-1)}{2 - (-1)} \]

1. **Calculate \(f(2)\) and \(f(-1)\)**:
\[ f(2) = 6, \quad f(-1) = 1 \]

2. **Calculate the right-hand side**:
\[ \frac{f(2) - f(-1)}{2 - (-1)} = \frac{6 - 1}{2 + 1} = \frac{5}{3} \]

3. **Set \(f'(c)\) equal to \(\frac{5}{3}\)**:
- For \(c < 1\):
\[ 3c^2 = \frac{5}{3} \implies c^2 = \frac{5}{9} \implies c = \pm \frac{\sqrt{5}}{3} \]
Since \(c\) must be in \([-1, 1)\), we take \(c = \frac{\sqrt{5}}{3}\).

- For \(c > 1\):
\[ f'(c) = 3 \text{ (which is constant)} \]
This does not equal \(\frac{5}{3}\).

### Conclusion

Thus, we have verified that the function satisfies the conditions of LMVT on the interval \([-1, 2]\) and found that \(c = \frac{\sqrt{5}}{3}\) is a point in the interval where the theorem holds true.

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