To determine which function satisfies the conditions of Rolle's theorem, we need to verify the three main criteria:
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 values at the endpoints must be equal, i.e., f(a) = f(b).
Let's analyze the provided options step by step.
### Step 1: Analyze Option 1
**Function:** \( f(x) = \log(x^2 + ab) - \log(a + b) - \log(x) \)
**Interval:** [a, b]
**Check continuity:**
- The logarithmic function is continuous wherever its argument is positive.
- Since \( x^2 + ab > 0 \) for all x in [a, b] (assuming a and b are positive), \( f(x) \) is continuous on [a, b].
**Check differentiability:**
- The function is differentiable wherever it is continuous, which is on (a, b) since logarithmic functions are differentiable in their domain.
**Check endpoint values:**
- Calculate \( f(a) \):
\[
f(a) = \log(a^2 + ab) - \log(a + b) - \log(a)
\]
Simplifying, we get:
\[
f(a) = \log\left(\frac{a^2 + ab}{(a + b)a}\right)
\]
- Calculate \( f(b) \):
\[
f(b) = \log(b^2 + ab) - \log(a + b) - \log(b)
\]
Simplifying, we get:
\[
f(b) = \log\left(\frac{b^2 + ab}{(a + b)b}\right)
\]
- We need to check if \( f(a) = f(b) \):
After simplification, we find that \( f(a) = f(b) \) holds true.
### Conclusion for Option 1:
Since all three conditions of Rolle's theorem are satisfied, **Option 1 is valid.**
### Step 2: Analyze Option 2
**Function:** \( f(x) = (x - 1)(2x - 3) \)
**Interval:** [1, 3]
**Check endpoint values:**
- Calculate \( f(1) \):
\[
f(1) = (1 - 1)(2 \cdot 1 - 3) = 0
\]
- Calculate \( f(3) \):
\[
f(3) = (3 - 1)(2 \cdot 3 - 3) = 2 \cdot 3 = 6
\]
Since \( f(1) \neq f(3) \), **Option 2 does not satisfy Rolle's theorem.**
### Step 3: Analyze Option 3
**Function:** \( f(x) = 2 + (x - 1)^{2/3} \)
**Interval:** [0, 2]
**Check endpoint values:**
- Calculate \( f(0) \):
\[
f(0) = 2 + (-1)^{2/3} = 2 + 1 = 3
\]
- Calculate \( f(2) \):
\[
f(2) = 2 + (2 - 1)^{2/3} = 2 + 1 = 3
\]
Since \( f(0) = f(2) \), we need to check continuity and differentiability:
- The function is continuous and differentiable on (0, 2).
### Conclusion for Option 3:
Since all conditions are satisfied, **Option 3 is valid.**
### Step 4: Analyze Option 4
**Function:** \( f(x) = \cos\left(\frac{1}{x}\right) \)
**Interval:** [-1, 1]
**Check differentiability:**
- At \( x = 0 \), the function is not defined, hence not differentiable.
### Conclusion for Option 4:
Since the function is not differentiable at \( x = 0 \), **Option 4 does not satisfy Rolle's theorem.**
### Final Answer:
The function in which Rolle's theorem is verified is **Option 1**.
