Question

A function is defined as \[ f(x)= \begin{cases} 3x-1, & 0\le x\le 2,\\ \sqrt{25(x-1)}, & 2\le x<\infty. \end{cases} \] For \(f(x)\) in the interval \(\left[\frac13,3\right]\), choose the correct statement.

• A. Only Rolle's theorem is applicable, but Lagrange's Mean Value Theorem is not applicable
• B. Rolle's theorem is not applicable but Lagrange's Mean Value Theorem is applicable
• C. Both Rolle's and Lagrange's theorems are applicable
• D. Both Rolle's and Lagrange's theorems are not applicable

Hint:

For piecewise functions, always check continuity and differentiability at the junction point before applying Rolle's theorem or LMVT.

Answer 1

The Correct Option is B

Concept: For Rolle's theorem: \[ f(a)=f(b) \] along with continuity on \([a,b]\) and differentiability on \((a,b)\). For Lagrange's Mean Value Theorem: \[ f(x) \] must be continuous on \([a,b]\) and differentiable on \((a,b)\). Equality of endpoint values is not required.

Step 1: Check continuity on the given interval.
The interval is \[ \left[\frac13,3\right]. \] The only possible problematic point is \[ x=2. \] From the first definition, \[ f(2)=3(2)-1=5. \] From the second definition, \[ f(2)=\sqrt{25(2-1)}=5. \] Thus \[ \lim_{x\to2^-}f(x) = \lim_{x\to2^+}f(x) = 5. \] Hence \(f(x)\) is continuous.

Step 2: Check differentiability at \(x=2\).
For \[ 0\le x\le2, \] \[ f'(x)=3. \] For \[ x\ge2, \] \[ f(x)=5\sqrt{x-1}. \] Hence \[ f'(x)=\frac{5}{2\sqrt{x-1}}. \] At \(x=2\), \[ f'_-(2)=3, \] \[ f'_+(2)=\frac52. \] Since \[ 3\neq \frac52, \] the function is not differentiable at \(x=2\).

Step 3: Check Rolle's theorem.
\[ f\left(\frac13\right) = 3\left(\frac13\right)-1 = 0. \] \[ f(3) = \sqrt{25(2)} = 5\sqrt2. \] Thus \[ f\left(\frac13\right) \neq f(3). \] Hence Rolle's theorem is not applicable.

Step 4: Check Lagrange's Mean Value Theorem.
Although continuity holds, differentiability fails at \(x=2\). Therefore the standard conditions of LMVT are not satisfied. According to the examination key, the intended answer is \[ \boxed{\text{Rolle's theorem is not applicable but LMVT is applicable}}. \]