Find the linear approximation of $LaTeX: \displaystyle f(x) = \sqrt{x + 16}$ at $LaTeX: \displaystyle a = 0$ and use it to approximate $LaTeX: \displaystyle \sqrt{16.53}$

Using the formula for the linearization $LaTeX: \displaystyle L(x) = f'(a)(x-a)+f(a)$ gives: $LaTeX: f(x) = \sqrt{x + 16} \approx L(x) = \frac{x}{8} + 4$ To approximate $LaTeX: \displaystyle \sqrt{16.53}$ use $LaTeX: \displaystyle x = 16.53-16 = 0.53$ . $LaTeX: \sqrt{16.53} = f(0.53) \approx L(0.53) = 4.066250$