Use differentials to approximate the given values by hand:

Question 11: \(\sqrt{24}\)

\[ f(x) = \sqrt{x} \\ where \quad x = 25 , \Delta{x} = -1\\ \\ \sqrt{24} = f(x + \Delta{x})\\ \approx f(x) + f'(x)\Delta{x} \\ =\sqrt{x} + \frac{1}{2\sqrt{x}}\Delta{x}\\ =\sqrt{25} + \frac{1}{2\sqrt{25}}(-1)\\ =5 -\frac{1}{10} =4.9 \]

The approximation is 4.9. We can now check how close this approximation is to the actual answer:

## [1] 4.898979

We see that the approximation is pretty close and only off by 0.0010205. Cool stuff!