Use linear approximation to estimate the following quantity. Choose a value of a to produce a small error. \[ \sqrt{131} \] $\sqrt{131} \approx \square$ (Round to three decimal places as needed.)

Step 1 ：Given the function \(f(x) = \sqrt{x}\) and a point \(a = 121\) where we know the exact value, we can use linear approximation to estimate \(\sqrt{131}\).

Step 2 ：The formula for linear approximation is \(L(x) = f(a) + f'(a)(x - a)\).

Step 3 ：First, we need to find the derivative of \(f(x) = \sqrt{x}\). The derivative is \(f'(x) = \frac{1}{2\sqrt{x}}\).

Step 4 ：So, \(f'(121) = \frac{1}{2\sqrt{121}} = \frac{1}{2\times 11} = \frac{1}{22}\).

Step 5 ：Now we can plug into the linear approximation formula: \(L(x) = f(121) + f'(121)(x - 121) = 11 + \frac{1}{22}(x - 121)\).

Step 6 ：We want to estimate \(\sqrt{131}\), so we plug in \(x = 131\): \(L(131) = 11 + \frac{1}{22}(131 - 121) = 11 + \frac{1}{22}\times 10 = 11 + 0.4545...\).

Step 7 ：So, \(\sqrt{131}\) is approximately 11.4545. Rounding to three decimal places, we get \(\sqrt{131} \approx 11.455\).

Step 8 ：This is a linear approximation, so there will be some error, but it should be relatively small.

Step 9 ：\(\boxed{\sqrt{131} \approx 11.455}\)