### Quiz 3.11:

Question:

Find the linearization of $$f(x)=\sqrt{x}$$ at $$x=4$$ and then use it to find an approximation of $$\sqrt{5}$$

Solution:

$$f^{\prime} (x)=\frac{1}{2\sqrt{x}}$$

$$f^{\prime} (4)=\frac{1}{2\sqrt{4}} = \frac{1}{4}$$

$$f (4)=\sqrt{4}=2$$

$$L(x) = 2 + \frac{1}{4} (x-4)=\frac{1}{4}x+1$$

$$\sqrt{5}=f(5) \approx L(5) = \frac{1}{4}(5)+1 = \frac{9}{4}=2.25$$