Let exp (x) denotes the exponetial function `e^(x)` . If f(x) = exp`(x^(1//x),x gt0` then the minimum value of f in the interval [2,5] is

Options

A. `exp(e^((1)/(e)))`
B. `exp(2^((1)/(2)))`
C. `exp(5^((1)/(5)))`
D. `exp(e^((2)/(e)))`

Answer

To find the minimum value of the function \( f(x) = e^{(x^{1/x})} \) in the interval \([2, 5]\), we will proceed step by step.

### Step 1: Define the function
We start with the function:
\[ f(x) = e^{(x^{1/x})} \]
To find the minimum value, we first need to analyze the exponent \( h(x) = x^{1/x} \).

### Step 2: Differentiate \( h(x) \)
To find the critical points of \( h(x) \), we will differentiate it. First, we take the natural logarithm:
\[ \ln(h(x)) = \frac{1}{x} \ln(x) \]
Now, differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln(h(x))) = \frac{d}{dx}\left(\frac{\ln(x)}{x}\right) \]
Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{\ln(x)}{x}\right) = \frac{x \cdot \frac{1}{x} - \ln(x)}{x^2} = \frac{1 - \ln(x)}{x^2} \]
Thus, we have:
\[ \frac{h'(x)}{h(x)} = \frac{1 - \ln(x)}{x^2} \]
This gives us:
\[ h'(x) = h(x) \cdot \frac{1 - \ln(x)}{x^2} \]

### Step 3: Find critical points
Set \( h'(x) = 0 \):
\[ h(x) \cdot \frac{1 - \ln(x)}{x^2} = 0 \]
Since \( h(x) \) is never zero for \( x > 0 \), we set:
\[ 1 - \ln(x) = 0 \implies \ln(x) = 1 \implies x = e \]

### Step 4: Evaluate \( h(x) \) at critical points and endpoints
Now we will evaluate \( h(x) \) at the endpoints of the interval \([2, 5]\) and at the critical point \( x = e \).

1. **At \( x = 2 \)**:
\[ h(2) = 2^{1/2} = \sqrt{2} \approx 1.414 \]

2. **At \( x = 5 \)**:
\[ h(5) = 5^{1/5} \approx 1.379 \]

3. **At \( x = e \)** (approximately 2.718):
\[ h(e) = e^{1/e} \approx 1.445 \]

### Step 5: Compare values
Now we compare the values of \( h(x) \):
- \( h(2) \approx 1.414 \)
- \( h(5) \approx 1.379 \)
- \( h(e) \approx 1.445 \)

The minimum value of \( h(x) \) in the interval \([2, 5]\) occurs at \( x = 5 \).

### Step 6: Find the minimum value of \( f(x) \)
Since \( f(x) = e^{h(x)} \), we need to find:
\[ f(5) = e^{h(5)} = e^{5^{1/5}} \approx e^{1.379} \]

### Conclusion
Thus, the minimum value of \( f(x) \) in the interval \([2, 5]\) is:
\[ f(5) = e^{5^{1/5}} \approx e^{1.379} \]