Calculate the absolute maximum and absolute minimum value of the function `f(x)=(x+1)/(sqrt(x^(2)+1)), 0lexle2.`

Answer

To find the absolute maximum and minimum values of the function \( f(x) = \frac{x+1}{\sqrt{x^2 + 1}} \) on the interval \( [0, 2] \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating \( f(x) \). The function is in the form of \( \frac{u}{v} \), where \( u = x + 1 \) and \( v = \sqrt{x^2 + 1} \).

Using the quotient rule:
\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
where \( u' = 1 \) and \( v' = \frac{x}{\sqrt{x^2 + 1}} \).

Thus, we have:
\[ f'(x) = \frac{\sqrt{x^2 + 1} \cdot 1 - (x + 1) \cdot \frac{x}{\sqrt{x^2 + 1}}}{x^2 + 1} \]

### Step 2: Simplify the derivative
Now we simplify the expression:
\[ f'(x) = \frac{\sqrt{x^2 + 1} - \frac{x(x + 1)}{\sqrt{x^2 + 1}}}{x^2 + 1} \]
Multiplying through by \( \sqrt{x^2 + 1} \):
\[ f'(x) = \frac{(x^2 + 1) - x(x + 1)}{(x^2 + 1)\sqrt{x^2 + 1}} \]
This simplifies to:
\[ f'(x) = \frac{1 - x^2}{(x^2 + 1)\sqrt{x^2 + 1}} \]

### Step 3: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = 1 \]

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical point \( x = 1 \) and the endpoints \( x = 0 \) and \( x = 2 \).

1. **At \( x = 0 \)**:
\[ f(0) = \frac{0 + 1}{\sqrt{0^2 + 1}} = \frac{1}{1} = 1 \]

2. **At \( x = 1 \)**:
\[ f(1) = \frac{1 + 1}{\sqrt{1^2 + 1}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414 \]

3. **At \( x = 2 \)**:
\[ f(2) = \frac{2 + 1}{\sqrt{2^2 + 1}} = \frac{3}{\sqrt{5}} \approx 1.341 \]

### Step 5: Compare the values
Now we compare the values obtained:
- \( f(0) = 1 \)
- \( f(1) = \sqrt{2} \approx 1.414 \)
- \( f(2) = \frac{3}{\sqrt{5}} \approx 1.341 \)

### Conclusion
The absolute maximum value of \( f(x) \) on the interval \( [0, 2] \) is \( \sqrt{2} \) at \( x = 1 \), and the absolute minimum value is \( 1 \) at \( x = 0 \).

### Final Answer
- Absolute Maximum: \( \sqrt{2} \)
- Absolute Minimum: \( 1 \)