Use linear approximation of the function $f$ given by $f(x) =\sqrt{16-x}$ at the point x = 0 to find an approximation of $\sqrt{15}$ by a rational number (i.e. fraction).
What I have so far:
$$L = \sqrt{16-0} - \frac{1}{2\sqrt{16-0}} \cdot \sqrt{15}\Rightarrow L = 4 - \frac{1}{8}\sqrt{15}$$
I don't know whether that's correct, and suppose that it is correct, how do I turn this into a fraction? Or is the answer simply: $$\frac{32-\sqrt{15}}{8}\text{ ?}$$
To summarise the comments:
You found that $\sqrt{16-x} \approx 4-\dfrac{x}{8}$ near $x=0$.
So letting $x=1$ suggests $\sqrt{16-1} \approx 4-\dfrac{1}{8}$, i.e. $\sqrt{15} \approx \dfrac{31}{8}$.
Indeed they are quite close as $\sqrt{15} \approx 3.873$ while $\frac{31}{8} = 3.875$.
In fact what you have done is use the tangent at one point on a parabola to approximate the value at another point nearby.