If `f(x) = 2x^(2)+5x+2`, then find the approximate value of `f(2.01)`.

Answer

To find the approximate value of \( f(2.01) \) for the function \( f(x) = 2x^2 + 5x + 2 \), we can use the concept of derivatives to estimate the change in the function value when \( x \) changes slightly from 2 to 2.01.

### Step-by-Step Solution:

1. **Identify the function**:
\[ f(x) = 2x^2 + 5x + 2 \]

2. **Calculate \( f(2) \)**:
\[ f(2) = 2(2)^2 + 5(2) + 2 \]
\[ = 2 \cdot 4 + 10 + 2 = 8 + 10 + 2 = 20 \]

3. **Determine \( \Delta x \)**:
\[ \Delta x = 2.01 - 2 = 0.01 \]

4. **Find the derivative \( f'(x) \)**:
\[ f'(x) = \frac{d}{dx}(2x^2 + 5x + 2) = 4x + 5 \]

5. **Evaluate the derivative at \( x = 2 \)**:
\[ f'(2) = 4(2) + 5 = 8 + 5 = 13 \]

6. **Calculate \( \Delta y \)** (the change in \( f \)):
\[ \Delta y \approx f'(2) \cdot \Delta x = 13 \cdot 0.01 = 0.13 \]

7. **Estimate \( f(2.01) \)**:
\[ f(2.01) \approx f(2) + \Delta y = 20 + 0.13 = 20.13 \]

### Final Answer:
\[ f(2.01) \approx 20.13 \]