Which topic is this question from? (Root-finding) 
Prove that if $x=a$ is an approximation to one root of the equation $f(x)=0$, then $x=a-\frac{f(a)}{f'(a)}$ is a closer approximation.

How to solve this question? Is this asking us to prove Newton-Raphson? 
 A: Use the quadratic Taylor formula, i.e., linear expansion with quadratic error term,
$$
f(a+h)=f(a)+f'(a)h+\frac12 f''(a+\theta h)h^2
$$
with $h=-\frac{f(a)}{f'(a)}$. Note that the constant and linear term cancel for the Newton update so that you get $f(x)=O(f(a)^2)$. For $x$ to be better than $a$ you need that the constant in this asymptotic estimate is small enough resp. that $a$ is close enough to the actual root.
You will see that the statement is not always true, you need certain relations between the function value $f(a)$ and bounds for the first and second derivative.
A: It needs some more careful statement in order to become subject to proof.
It is not always true that a Newton iteration produces a closer approximation to a root, though there is a basin of attraction around a simple root of a smooth (e.g. twice continuously differentiable) function wherein the iterations will converge to that root.
The analysis of quadratic convergence for Newton-Raphson can be carried out using Taylor's theorem for such functions.
Essentially showing that the iteration yields an approximation closer to the root depends on having the current "guess" close enough to make a term that involves $f''(x)/f'(x)$ times the square of the current error smaller than the current error.  Therefore we want $|f''(x)|$ bounded above and $|f'(x)|$ bounded below (away from zero), and the current approximation needs to be accurate enough to make the errors shrink.
Let me know if the Wikipedia presentation does not give you sufficient detail.  A proof can be found in most advanced numerical analysis texts, and sometimes in mathematical analysis texts as it makes a good illustration of Taylor's theorem.
