The real number x when added to its inverse gives the minimum value of the sum at x equal to

Options

A. `-2`
B. `-1`
C. 1
D. 2

Answer

To find the real number \( x \) such that when added to its inverse gives the minimum value of the sum, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x + \frac{1}{x} \).

### Step 2: Find the derivative
To find the minimum value, we need to find the derivative of \( f(x) \):
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 3: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 1 - \frac{1}{x^2} = 0 \]

### Step 4: Solve for \( x \)
Rearranging gives:
\[ \frac{1}{x^2} = 1 \implies x^2 = 1 \]
Thus, \( x = 1 \) or \( x = -1 \).

### Step 5: Determine the nature of critical points
To determine whether these points are minima or maxima, we can use the second derivative test or analyze the first derivative:
\[ f''(x) = \frac{2}{x^3} \]
- For \( x = 1 \):
\[ f''(1) = 2 > 0 \quad \text{(indicating a local minimum)} \]
- For \( x = -1 \):
\[ f''(-1) = -2 < 0 \quad \text{(indicating a local maximum)} \]

### Step 6: Find the minimum value
Now we can find the minimum value of \( f(x) \) at \( x = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

### Conclusion
The real number \( x \) when added to its inverse gives the minimum value of the sum at \( x = 1 \), and the minimum value of the sum is \( 2 \).

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