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