The greatest of the numbers `2^(1/2), 3^(1/3), 4^(1/4), 5^(1/5), 6^(1/6)` and `7^(1/7)` is

Options

A. `2 ^(1//2)`
B. `3 ^(1//3)`
C. `7 ^(1//7)`
D. `6 ^(1//6)`

Answer

To find the greatest number among \(2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6},\) and \(7^{1/7}\), we can analyze the function \(y = x^{1/x}\). We will use calculus to find the maximum value of this function.

### Step 1: Define the function
Let \(y = x^{1/x}\).

### Step 2: Take the natural logarithm
To simplify the differentiation, we take the natural logarithm of both sides:
\[ \ln y = \frac{1}{x} \ln x \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \(x\):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\ln x}{x} \right) \]
Using the quotient rule on the right side:
\[ \frac{d}{dx} \left( \frac{\ln x}{x} \right) = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \]

### Step 4: Set the derivative to zero
To find the critical points, we set \(\frac{dy}{dx} = 0\):
\[ \frac{1 - \ln x}{x^2} = 0 \]
This implies:
\[ 1 - \ln x = 0 \implies \ln x = 1 \implies x = e \]
where \(e \approx 2.718\).

### Step 5: Evaluate the function at integer values
Since we are interested in the integers from 2 to 7, we need to evaluate \(y\) at these points:
- For \(x = 2\):
\[ y = 2^{1/2} = \sqrt{2} \approx 1.414 \]
- For \(x = 3\):
\[ y = 3^{1/3} \approx 1.442 \]
- For \(x = 4\):
\[ y = 4^{1/4} = \sqrt{2} \approx 1.414 \]
- For \(x = 5\):
\[ y = 5^{1/5} \approx 1.379 \]
- For \(x = 6\):
\[ y = 6^{1/6} \approx 1.348 \]
- For \(x = 7\):
\[ y = 7^{1/7} \approx 1.349 \]

### Step 6: Compare the values
Now, we compare the values:
- \(2^{1/2} \approx 1.414\)
- \(3^{1/3} \approx 1.442\)
- \(4^{1/4} \approx 1.414\)
- \(5^{1/5} \approx 1.379\)
- \(6^{1/6} \approx 1.348\)
- \(7^{1/7} \approx 1.349\)

The greatest value is \(3^{1/3} \approx 1.442\).

### Final Answer
Thus, the greatest of the numbers \(2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6},\) and \(7^{1/7}\) is:
\[ \boxed{3^{1/3}} \]