How can I figure out this question?
Use linear approximation i.e. tangent line to approximate square root of 49.1 as follows,
Let f(x0 = square root of x and the equation of the tangent line to f(x) can be written in the form x=49 in the form y=mx+b.
where m is:
and b is:
Using this we find our approximation for square root of 49.1 is...

Question

How can I figure out this question? 
Use linear approximation i.e. tangent line to approximate square root of 49.1 as follows, 
Let f(x0 = square root of x and the equation of the tangent line to f(x) can be written in the form x=49 in the form y=mx+b.
where m is:
and b is:
Using this we find our approximation for square root of 49.1 is...

anonymous · Answer

The linear approximation to a function $f(x)$ at a point $c$ is
$$f(x)\approx f'(c)(x-c)+f(c)$$
To approximate $\sqrt{49.1}$, you would set $x=49.1$ and $c$ to some value whose square root is easy to find and close to the given $x$. In this case, $c=49$.

$$f(x)=\sqrt{x}~~\Rightarrow~~f'(x)=\frac{1}{2\sqrt x}\
f(49.1)\approx f'(49)(49.1-49)+\sqrt{49}\
~~~~~~~~~~~~=\frac{1}{2\sqrt{49}}(0.1)+7\
~~~~~~~~~~~~=\cdots$$