The minimum value of the fuction f(x) given by `f(x)=(x^m)/(m)+(x^(-n))/(n) " where " 1/m+1/n =1" and " m gt 1` is

Options

A. 1
B. 0
C. 2
D. non of these

Explanation

To find the minimum value of the function \( f(x) = \frac{x^m}{m} + \frac{x^{-n}}{n} \) given the condition \( \frac{1}{m} + \frac{1}{n} = 1 \) and \( m > 1 \), we can follow these steps:

### Step 1: Understand the relationship between \( m \) and \( n \)
From the equation \( \frac{1}{m} + \frac{1}{n} = 1 \), we can rearrange it to find \( n \) in terms of \( m \):
\[ \frac{1}{n} = 1 - \frac{1}{m} \implies n = \frac{m}{m-1} \]
This shows that \( n > 1 \) since \( m > 1 \).

### Step 2: Differentiate the function
To find the minimum value, we need to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}\left(\frac{x^m}{m}\right) + \frac{d}{dx}\left(\frac{x^{-n}}{n}\right) \]
Calculating the derivatives:
\[ f'(x) = \frac{m x^{m-1}}{m} - \frac{n x^{-n-1}}{n} = x^{m-1} - x^{-n-1} \]

### Step 3: Set the derivative to zero
To find critical points, set \( f'(x) = 0 \):
\[ x^{m-1} - x^{-n-1} = 0 \implies x^{m-1} = x^{-n-1} \]
Multiplying both sides by \( x^{n+1} \) (assuming \( x > 0 \)):
\[ x^{m+n} = 1 \implies x = 1 \]

### Step 4: Verify if it is a minimum
To confirm that this critical point is a minimum, we need to check the second derivative:
\[ f''(x) = (m-1)x^{m-2} + (n+1)x^{-n-2} \]
Evaluating the second derivative at \( x = 1 \):
\[ f''(1) = (m-1) + (n+1) = m + n > 0 \]
Since \( f''(1) > 0 \), \( x = 1 \) is indeed a minimum point.

### Step 5: Calculate the minimum value
Now, substitute \( x = 1 \) into the original function to find the minimum value:
\[ f(1) = \frac{1^m}{m} + \frac{1^{-n}}{n} = \frac{1}{m} + \frac{1}{n} \]
Using the relationship \( \frac{1}{m} + \frac{1}{n} = 1 \):
\[ f(1) = 1 \]

### Conclusion
Thus, the minimum value of the function \( f(x) \) is:
\[ \boxed{1} \]