Questions: Find the local linear approximation of the function f(x) = sqrt(13+x) at x0 = 12, and use it to approximate sqrt(24.9) and sqrt(25.1). (a) f(x) = sqrt(13+x) ≈ (b) sqrt(24.9) ≈ (c) sqrt(25.1) ≈ For parts (b) and (c), you should enter your answer as a fraction. If you enter a decimal, make sure that it is correct to at least six decimal places.

Solution Steps

To find the local linear approximation of the function $f(x) = \sqrt{13+x}$ at $x_0 = 12$, we first need to compute the derivative of the function, $f'(x)$. The linear approximation at $x_0$ is given by $L(x) = f(x_0) + f'(x_0)(x - x_0)$. We then use this linear approximation to estimate $\sqrt{24.9}$ and $\sqrt{25.1}$ by substituting $x = 11.9$ and $x = 12.1$ respectively, since $\sqrt{24.9} = \sqrt{13 + 11.9}$ and $\sqrt{25.1} = \sqrt{13 + 12.1}$.

To find the local linear approximation of the function $f(x) = \sqrt{13+x}$ at $x_0 = 12$, we first compute the derivative: $$f'(x) = \frac{1}{2\sqrt{13+x}}$$ Evaluating this at $x_0 = 12$: $$f'(12) = \frac{1}{2\sqrt{25}} = \frac{1}{10}$$

The linear approximation $L(x)$ at $x_0 = 12$ is given by: $$L(x) = f(12) + f'(12)(x - 12)$$ Calculating $f(12)$: $$f(12) = \sqrt{25} = 5$$ Thus, the linear approximation becomes: $$L(x) = 5 + \frac{1}{10}(x - 12) = \frac{1}{10}x + \frac{19}{5}$$

To approximate $\sqrt{24.9}$ and $\sqrt{25.1}$, we substitute $x = 11.9$ and $x = 12.1$ into the linear approximation $L(x)$.

For $\sqrt{24.9}$: $$L(11.9) = \frac{1}{10}(11.9) + \frac{19}{5} = 4.990$$

For $\sqrt{25.1}$: $$L(12.1) = \frac{1}{10}(12.1) + \frac{19}{5} = 5.010$$

Final Answer