For `xge 0` , the smallest value of the function `f(x)=(4x^2+8x+13)/(6(1+x))`, is ________.

Explanation

To find the smallest value of the function \( f(x) = \frac{4x^2 + 8x + 13}{6(1+x)} \) for \( x \geq 0 \), we can follow these steps:

### Step 1: Simplify the function
We start with the function:
\[ f(x) = \frac{4x^2 + 8x + 13}{6(1+x)} \]

### Step 2: Rewrite the numerator
The numerator \( 4x^2 + 8x + 13 \) can be rewritten as:
\[ 4x^2 + 8x + 13 = 4(x^2 + 2x) + 13 \]
Completing the square for \( x^2 + 2x \):
\[ x^2 + 2x = (x+1)^2 - 1 \]
Thus,
\[ 4(x^2 + 2x) + 13 = 4((x+1)^2 - 1) + 13 = 4(x+1)^2 - 4 + 13 = 4(x+1)^2 + 9 \]

### Step 3: Substitute back into the function
Now substituting back, we have:
\[ f(x) = \frac{4(x+1)^2 + 9}{6(1+x)} = \frac{4(x+1)^2}{6(1+x)} + \frac{9}{6(1+x)} \]
This simplifies to:
\[ f(x) = \frac{2}{3}(x+1) + \frac{3}{2(x+1)} \]

### Step 4: Use AM-GM inequality
To find the minimum value of \( f(x) \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. We set:
\[ A = \frac{2}{3}(x+1) \quad \text{and} \quad B = \frac{3}{2(x+1)} \]
According to AM-GM:
\[ \frac{A + B}{2} \geq \sqrt{AB} \]

### Step 5: Calculate \( AB \)
Calculating \( AB \):
\[ AB = \left(\frac{2}{3}(x+1)\right) \left(\frac{3}{2(x+1)}\right) = 1 \]

### Step 6: Apply AM-GM
Thus, we have:
\[ \frac{A + B}{2} \geq 1 \implies A + B \geq 2 \]
This means:
\[ f(x) \geq 2 \]

### Step 7: Find when equality holds
Equality in AM-GM holds when \( A = B \):
\[ \frac{2}{3}(x+1) = \frac{3}{2(x+1)} \]
Cross-multiplying gives:
\[ 4(x+1)^2 = 9 \implies (x+1)^2 = \frac{9}{4} \implies x+1 = \frac{3}{2} \implies x = \frac{1}{2} \]

### Conclusion
Thus, the smallest value of the function \( f(x) \) is:
\[ \boxed{2} \]