Which of the following functions satisfies the conditions of Rolle's theorem?

Question

Accepted Answer

To determine which function satisfies the conditions of Rolle's theorem, we need to verify that the function meets the following criteria on the given interval:

1. The function must be continuous on the closed interval $[a, b]$.
2. The function must be differentiable on the open interval $(a, b)$.
3. The function must have equal values at the endpoints, i.e., $f(a) = f(b)$.

Let's analyze each option:

**Option 1: $f(x) = \frac{x-1}{\sqrt{x}}; x \in [0,1]$**

- **Continuity:** The function $f(x) = \frac{x-1}{\sqrt{x}}$ is not continuous at $x = 0$ because $\sqrt{x}$ in the denominator becomes zero, making the function undefined at $x = 0$.
- **Differentiability:** Since the function is not continuous at $x = 0$, it cannot be differentiable on the interval $(0, 1)$.
- **Endpoint Values:** The function is not defined at $x = 0$, so we cannot evaluate $f(0)$.

Since the function is not continuous on $[0, 1]$, it does not satisfy Rolle's theorem.

**Option 2: $f(x) = \sqrt{x}(x-1); x \in [0,1]$**

- **Continuity:** The function $f(x) = \sqrt{x}(x-1)$ is continuous on $[0, 1]$ because both $\sqrt{x}$ and $x-1$ are continuous functions, and the product of continuous functions is continuous.
- **Differentiability:** The function is differentiable on $(0, 1)$ because both $\sqrt{x}$ and $x-1$ are differentiable, and the product rule can be applied.
- **Endpoint Values:** At $x = 0$, $f(0) = \sqrt{0}(0-1) = 0$. At $x = 1$, $f(1) = \sqrt{1}(1-1) = 0$. Thus, $f(0) = f(1)$.

Since all conditions of Rolle's theorem are satisfied, Option 2 is correct.

**Option 3: $f(x) = \frac{\tan x}{x}; x \in [0, \pi]$**

- **Continuity:** The function is not continuous at $x = 0$ because $\tan x$ is undefined at $x = 0$ and $\frac{\tan x}{x}$ is undefined at $x = 0$.
- **Differentiability:** Since the function is not continuous at $x = 0$, it cannot be differentiable on the interval $(0, \pi)$.
- **Endpoint Values:** The function is not defined at $x = 0$, so we cannot evaluate $f(0)$.

Since the function is not continuous on $[0, \pi]$, it does not satisfy Rolle's theorem.

**Option 4: $f(x) = \sin\frac{1}{x}; x \in [-\frac{1}{\pi}, \frac{1}{\pi}]$**

- **Continuity:** The function is not continuous at $x = 0$ because $\frac{1}{x}$ becomes undefined, making $\sin\frac{1}{x}$ undefined at $x = 0$.
- **Differentiability:** Since the function is not continuous at $x = 0$, it cannot be differentiable on the interval $(-1/\pi, 1/\pi)$.
- **Endpoint Values:** The function is not defined at $x = 0$, so we cannot evaluate $f(-1/\pi)$ and $f(1/\pi)$.

Since the function is not continuous on $[-1/\pi, 1/\pi]$, it does not satisfy Rolle's theorem.

In conclusion, Option 2 is the only function that satisfies all the conditions of Rolle's theorem on the given interval.