If `f(x)=(3x^(2)+1)/(sqrt(x^(4)+x^(2))) AA 0gtxlesqrt(2)`, then `f(x)` has minimum value `M` at `x=m`. Then `(m^(2) +1/(M^(2)))` is equal to

Explanation

To solve the problem step by step, we will analyze the function \( f(x) = \frac{3x^2 + 1}{\sqrt{x^4 + x^2}} \) for \( 0 < x \leq \sqrt{2} \) and find its minimum value \( M \) at \( x = m \). We will then compute \( m^2 + \frac{1}{M^2} \).

### Step 1: Simplify the function

We start with the function:
\[ f(x) = \frac{3x^2 + 1}{\sqrt{x^4 + x^2}} \]
We can factor the denominator:
\[ \sqrt{x^4 + x^2} = \sqrt{x^2(x^2 + 1)} = x\sqrt{x^2 + 1} \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = \frac{3x^2 + 1}{x\sqrt{x^2 + 1}} = \frac{3x^2}{x\sqrt{x^2 + 1}} + \frac{1}{x\sqrt{x^2 + 1}} = \frac{3x}{\sqrt{x^2 + 1}} + \frac{1}{x\sqrt{x^2 + 1}} \]

### Step 2: Find critical points

To find the minimum value, we will differentiate \( f(x) \) and set the derivative to zero. Let:
\[ f(x) = \frac{3x}{\sqrt{x^2 + 1}} + \frac{1}{x\sqrt{x^2 + 1}} \]
We will use the quotient rule and product rule to differentiate \( f(x) \).

### Step 3: Differentiate \( f(x) \)

Using the quotient rule:
\[ f'(x) = \frac{(3\sqrt{x^2 + 1} - \frac{3x^2}{\sqrt{x^2 + 1}}) + \left(-\frac{1}{x^2\sqrt{x^2 + 1}} + \frac{1}{x}\cdot\frac{x}{2\sqrt{x^2 + 1}}\right)}{(x^2 + 1)} \]
Setting \( f'(x) = 0 \) will give us the critical points.

### Step 4: Solve for critical points

After differentiating and simplifying, we will find the critical points \( x \) that satisfy \( f'(x) = 0 \).

### Step 5: Evaluate \( f(x) \) at critical points and endpoints

We will evaluate \( f(x) \) at the critical points found in Step 4 and at the endpoints \( x = 0 \) and \( x = \sqrt{2} \).

### Step 6: Determine minimum value \( M \) and corresponding \( m \)

From the evaluations, we will find the minimum value \( M \) and the corresponding \( m \).

### Step 7: Calculate \( m^2 + \frac{1}{M^2} \)

Finally, we will compute:
\[ m^2 + \frac{1}{M^2} \]