Find the minimum values of `f(x)=x^(2)+(1)/(x^(2)), x gt 0`

Answer

To find the minimum value of the function \( f(x) = x^2 + \frac{1}{x^2} \) for \( x > 0 \), we can follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{1}{x^2}\right) \]

Calculating the derivatives:

\[ f'(x) = 2x - \frac{2}{x^3} \]

### Step 2: Set the derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ 2x - \frac{2}{x^3} = 0 \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:

\[ 2x = \frac{2}{x^3} \]

Cross-multiplying yields:

\[ 2x^4 = 2 \]

Dividing both sides by 2:

\[ x^4 = 1 \]

Taking the fourth root gives:

\[ x = 1 \quad (\text{since } x > 0) \]

### Step 4: Determine if it is a minimum
Next, we need to check if this critical point is a minimum by using the second derivative test. We find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}\left(2x - \frac{2}{x^3}\right) \]

Calculating the derivative:

\[ f''(x) = 2 + \frac{6}{x^4} \]

### Step 5: Evaluate the second derivative at the critical point
Now we evaluate \( f''(1) \):

\[ f''(1) = 2 + \frac{6}{1^4} = 2 + 6 = 8 \]

Since \( f''(1) > 0 \), this indicates that \( f(x) \) has a local minimum at \( x = 1 \).

### Step 6: Calculate the minimum value
Finally, we find the minimum value of the function by substituting \( x = 1 \) back into the original function:

\[ f(1) = 1^2 + \frac{1}{1^2} = 1 + 1 = 2 \]

Thus, the minimum value of \( f(x) \) for \( x > 0 \) is:

\[ \boxed{2} \]