On the interval [0, 1] the function $x^{25}(1-x)^{75}$ takes its maximum value at the point:

Question

Accepted Answer

To determine why the function $ f(x) = x^{25}(1-x)^{75} $ takes its maximum value at $ x = \frac{1}{4} $ on the interval $[0, 1]$, we can use calculus to find the critical points and analyze the behavior of the function.

### Explanation for Option 4 ($ x = \frac{1}{4} $):

1. **Finding the Critical Points**:
   - The function $ f(x) = x^{25}(1-x)^{75} $ is a product of two terms, $ x^{25} $ and $ (1-x)^{75} $.
   - To find the critical points, we take the derivative of $ f(x) $ and set it to zero:
     $$
     f'(x) = \frac{d}{dx}[x^{25}(1-x)^{75}] = 25x^{24}(1-x)^{75} - 75x^{25}(1-x)^{74}
     $$
   - Setting $ f'(x) = 0 $ gives:
     $$
     25x^{24}(1-x)^{75} = 75x^{25}(1-x)^{74}
     $$
   - Simplifying, we get:
     $$
     25(1-x) = 75x \quad \Rightarrow \quad 25 = 100x \quad \Rightarrow \quad x = \frac{1}{4}
     $$

2. **Second Derivative Test**:
   - To confirm that $ x = \frac{1}{4} $ is a maximum, we use the second derivative test.
   - Compute the second derivative $ f''(x) $ and evaluate it at $ x = \frac{1}{4} $.
   - If $ f''(x) < 0 $ at $ x = \frac{1}{4} $, then $ f(x) $ has a local maximum at this point.

3. **Behavior at the Endpoints**:
   - At $ x = 0 $, $ f(0) = 0^{25}(1-0)^{75} = 0 $.
   - At $ x = 1 $, $ f(1) = 1^{25}(1-1)^{75} = 0 $.
   - Since $ f(x) $ is zero at both endpoints and positive within the interval, the maximum must occur at a critical point within $(0, 1)$.

### Explanation for Why Other Options are Incorrect:

1. **Option 1 ($ x = 0 $)**:
   - As calculated, $ f(0) = 0 $. The function value is zero at this endpoint, not a maximum.

2. **Option 2 ($ x = \frac{1}{3} $)**:
   - While $ x = \frac{1}{3} $ is within the interval, it is not a critical point derived from setting the derivative to zero. Evaluating $ f(x) $ at $ x = \frac{1}{3} $ does not yield the maximum value compared to $ x = \frac{1}{4} $.

3. **Option 3 ($ x = \frac{1}{2} $)**:
   - Similarly, $ x = \frac{1}{2} $ is not a critical point for this function. Evaluating $ f(x) $ at $ x = \frac{1}{2} $ results in a smaller value than at $ x = \frac{1}{4} $.

In conclusion, the function $ f(x) = x^{25}(1-x)^{75} $ achieves its maximum value at $ x = \frac{1}{4} $, making Option 4 the correct answer.