The given problem asks to use linear approximation for the function $f(x) = \sqrt{49 - x}$ at $x = 0$, and to approximate $f(1) = 4 \cdot \sqrt{3}$ by a rational number.

Step 1: Linear Approximation Formula

The formula for linear approximation is: $f(x) \approx f(a) + f'(a)(x - a)$ where $f'(a)$ is the derivative of $f(x)$ at $x = a$.

Step 2: Function $f(x) = \sqrt{49 - x}$

We are given $f(x) = \sqrt{49 - x}$. Let's calculate the derivative:

$f'(x) = \frac{d}{dx} \left( \sqrt{49 - x} \right) = \frac{1}{2\sqrt{49 - x}} \cdot (-1) = -\frac{1}{2\sqrt{49 - x}}$

Step 3: Evaluate at $x = 0$

Now we evaluate $f(x)$ and $f'(x)$ at $x = 0$:

$f(0) = \sqrt{49 - 0} = \sqrt{49} = 7$

$f'(0) = -\frac{1}{2\sqrt{49}} = -\frac{1}{2 \cdot 7} = -\frac{1}{14}$

Step 4: Linear Approximation for $f(1)$

Now, use the linear approximation formula at $x = 0$ to approximate $f(1)$:

$f(1) \approx f(0) + f'(0)(1 - 0) = 7 + \left( -\frac{1}{14} \right)(1) = 7 - \frac{1}{14} = \frac{98}{14} - \frac{1}{14} = \frac{97}{14}$

Thus, $f(1) \approx \frac{97}{14}$.

Step 5: Approximation of $4 \cdot \sqrt{3}$

We are asked to approximate $4 \cdot \sqrt{3}$ by a rational number. Using the approximation $\sqrt{3} \approx 1.732$, we get:

$4 \cdot \sqrt{3} \approx 4 \cdot 1.732 = 6.928$

Hence, the rational approximation is close to $\frac{97}{14}$, as calculated earlier.

Do you want to go over these steps in more detail or have any further questions?

Related Questions:

1. How can linear approximation be applied to other functions?
2. What are the limitations of linear approximation?
3. Why does linear approximation work well near $x = 0$?
4. How do we improve the accuracy of this approximation?
5. What is the difference between linear approximation and a Taylor series?

Tip:

Linear approximation is most accurate near the point where the function is approximated, so always check how far you are extrapolating!