If f(x) =`{:{(2x^2+2/x^2",",0ltabsxle2),(3",",x gt2):}`then

Options

A. (a) `x=1,-1` are the points of global minima
B. (b) x=1,-1 are the points of local minima
C. (c) x=0 is the point of local minima
D. (d) `x=0` is the point of local minimum

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} 2x^2 + \frac{2}{x^2} & \text{if } 0 < |x| \leq 2 \\ 3 & \text{if } x > 2 \end{cases} \]

### Step 1: Find the derivative of \( f(x) \) for \( 0 < |x| \leq 2 \)

We start by differentiating the function \( f(x) = 2x^2 + \frac{2}{x^2} \).

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

Using the power rule and the quotient rule, we get:

\[ f'(x) = 4x - \frac{4}{x^3} \]

### Step 2: Set the derivative equal to zero to find critical points

To find the critical points, we set \( f'(x) = 0 \):

\[ 4x - \frac{4}{x^3} = 0 \]

Multiplying through by \( x^3 \) (assuming \( x \neq 0 \)) gives:

\[ 4x^4 - 4 = 0 \implies x^4 = 1 \]

Taking the fourth root, we find:

\[ x = 1 \quad \text{and} \quad x = -1 \]

### Step 3: Determine the nature of the critical points using the second derivative

Next, we find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(4x - \frac{4}{x^3}) = 4 + \frac{12}{x^4} \]

Now we evaluate \( f''(x) \) at the critical points \( x = 1 \) and \( x = -1 \):

1. For \( x = 1 \):

\[ f''(1) = 4 + \frac{12}{1^4} = 4 + 12 = 16 > 0 \quad \text{(indicating a local minimum)} \]

2. For \( x = -1 \):

\[ f''(-1) = 4 + \frac{12}{(-1)^4} = 4 + 12 = 16 > 0 \quad \text{(indicating a local minimum)} \]

### Step 4: Evaluate the function at the critical points

Now we evaluate \( f(x) \) at the critical points:

1. \( f(1) = 2(1^2) + \frac{2}{1^2} = 2 + 2 = 4 \)
2. \( f(-1) = 2(-1)^2 + \frac{2}{(-1)^2} = 2 + 2 = 4 \)

### Step 5: Compare with the value of \( f(x) \) when \( x = 2 \)

Next, we check the value of \( f(2) \):

\[ f(2) = 2(2^2) + \frac{2}{2^2} = 8 + \frac{2}{4} = 8 + 0.5 = 8.5 \]

### Step 6: Determine the global minimum

Since \( f(x) = 3 \) for \( x > 2 \), we compare the local minima values:

- \( f(1) = 4 \)
- \( f(-1) = 4 \)
- \( f(2) = 8.5 \)
- \( f(x) = 3 \) for \( x > 2 \)

The minimum value of \( f(x) \) occurs at \( x > 2 \) where \( f(x) = 3 \).

### Conclusion

The local minima at \( x = 1 \) and \( x = -1 \) are both equal to 4, but the global minimum occurs at \( x > 2 \) where \( f(x) = 3 \).