Use linear approximation to estimate the following quantity. Choose a value of a to produce a small error. \[ \sqrt{150} \]

Step 1 ：Given function is \(f(x) = \sqrt{x}\) and we want to approximate it at \(x = 150\).

Step 2 ：We choose \(a = 144\) because the square root of 144 is a whole number (12), and it's close to 150.

Step 3 ：Compute \(f(a) = \sqrt{a} = \sqrt{144} = 12\).

Step 4 ：The derivative of \(f(x) = \sqrt{x}\) is \(f'(x) = \frac{1}{2\sqrt{x}}\). So, compute \(f'(a) = \frac{1}{2\sqrt{a}} = \frac{1}{2\sqrt{144}} = \frac{1}{24}\).

Step 5 ：Use the equation of the tangent line to approximate \(\sqrt{150}\): \(\sqrt{150} \approx f(a) + f'(a)(x - a) = 12 + \frac{1}{24}(150 - 144) = 12 + \frac{1}{24}\times6 = 12.25\).

Step 6 ：\(\boxed{12.25}\) is the linear approximation of \(\sqrt{150}\).

Step 7 ：The actual value of \(\sqrt{150}\) is approximately 12.2474. Our approximation is very close to the actual value, so we can conclude that our choice of \(a = 144\) produced a small error.