First Order Approximation of a $\sqrt{3.9}$ 
Use the first order approximation to determine $\sqrt{3.9}$. Use your calculator to determine the error in your approximation.

I am really confused by this question. There is no function so I don't know how can I use the Taylor approximation? Can I assume the $f(x) = \sqrt{x}$ ?
If I can assume that I can use the first order approximation but I am not sure whether I can do that assumption.
Note: Thank you all for your help. As it turns out I can assume $f(x) = \sqrt{x}$. I also found this YouTube video which I believe will help anyone who has the same problem: Approximation of Square Roots by Taylor Approximation
 A: The first order approximation means you use the Taylor series of $f(x) = \sqrt{x}$ up to the term of degree 1 (=first order). In this case it is very convenient to develop it around $x_0=4$, as it is close to $3.9$ and we already know that $\sqrt{4}=2$:
$$f(x_0+h)\approx f(x_0) + hf'(x_0)$$
In our case $x_0=4$ and $h=-0.1$ and with $f(x) = \sqrt{x}$ we get $f'(x) = \frac{1}{2\sqrt{x}}$ so
$$f(3.9) = f(x_0+h) \approx f(4) -0.1f'(4) = 2 -0.1\frac14 = 2-0.025 = 1.975$$
Actually $\sqrt{3.9} = 1.974841...$ (computed using a calculator) so our error is at about $1.6\cdot 10^{-4}$.
A: Since
$\sqrt{1+x}
\approx 1+\frac{x}{2}
$
for small $x$,
$\begin{array}\\
\sqrt{a^2+b}
&=a\sqrt{1+\frac{b}{a^2}}\\
&\approx a(1+\frac{b}{2a^2})\\
& a+\frac{b}{2a}\\
\end{array}
$
If
$a=2, b=-.1$,
$\sqrt{3.9}
\approx 2-\frac{.1}{2\cdot 2}
= 2-\frac1{40}
=2-.025
=1.975
$.
Note that
$(2-\frac1{40})^2
=4-4\cdot\frac1{40}+\frac1{1600}
=3.9+\frac1{1600}
$
and,
in general,
$(a+\frac{b}{2a})^2
=a^2+2a\frac{b}{2a}+\frac{b^2}{4a^2}
=a^2+b+\frac{b^2}{4a^2}
$.
