You have discovered a variant of the standard functional iteration or fixed point iteration. The standard iteration is $$x_{n+1} = f(x_n).$$ You are using $$x_{n+1} = g(x_n),$$ where $$g(x) = (x + f(x))/2.$$
EDIT: A recent question and its answer by @Lutz started a thought process which prompted my decision to revise this answer thoroughly. Briefly, it is possible to construct functional iterations which converge at least quadratically without recomputing derivatives as required for Newton's method. However, since it is necessary to evaluate the derivative at the fixed point, this technique is mostly limit to the construction of good problems for students.
Let $I \subset \mathbb{R}$. A function $h : I \rightarrow I$ is a contraction, if there exists $L \in [0,1)$, such that
$$\forall x, y \: : \: |h(x) - h(y)| \leq L | x-y|$$
If $I$ is compact and if $h : I \rightarrow I$ is a contraction, then $h$ has precisely one fixed point $z \in I$ and the functional iteration
$$ x_{n+1} = h(x_n)$$ converges to $z$ for any starting value $x_0 \in I$.
If $I = [a,b]$ and $h \in C^1(I)$ has a fixed point $z \in [a,b]$ and $|h'(z)| < 1$, then there exists compact interval $J \subseteq I$ such that the restriction of $h$ to $J$ is a contraction. This follows from the mean value theorem and the continuity of $h'$.
We can exploit the above mentioned property to design functional iterations which are convergent. Suppose $f$ has a fixed point $z$, but $|f'(z)|>1$. Then as stated by @Ian, the iteration driven by $f$ will not converge to $z$, but the function $g$ given by $$g(x) = \frac{f(x)-f'(z)x}{1-f'(z)}$$ satisfies
$$ g(z) = \frac{f(z)-f'(z)z}{1-f'(z)} = \frac{z-f'(z)z}{1-f'(z)} = \frac{1-f'(z)}{1-f'(z)}z= z$$
and $$ g'(x) = \frac{f'(x)-f'(z)}{1-f'(z)} \quad \Rightarrow \quad g'(z) = 0.$$ It follows that the functional iteration driven by $g$ will converge at least quadratically to $z$. In practice, it is not realistic to assume that the exact value of $f'(z)$ is available, but we can have linear convergence if we know a good approximation. Specifically, suppose $c \approx f'(z)$ and $c \not = 1$. Then we consider the function $h$ given by $$h(x) = \frac{f(x)-cx}{1-c}.$$ As before, we have $$h(z) = z$$ and $$h'(x) = \frac{f'(x)-c}{1-c}$$ which implies $$h'(z) = \frac{f'(z)-c}{1-c}.$$
Therefore, $|h'(z)| < 1$ provided that $|f'(z)-c| < |1-c|$.
The aforementioned question by @UmShmum is provides an excellent test case for these ideas. Consider the non-linear equation $x = f(x)$ where $$f(x) = 2 \sin(x).$$ It is clear that $z=0$ is a fixed point of $f$, but since $f'(x) = 2 \cos(x)$ implies $f'(0) = 2$, the standard functional iteration will not converge. However, if $g$ is given by
$$g(x) = \frac{f(x)-f'(0)x}{1-f'(0)} = 2(x-\sin(x)),$$ then
$g(0) = 0$ and $g'(x)=2(1-\cos(x)$ implies $g'(x) = 0$. It follows that the functional iteration given by $g$ will converge at least quadratically to $0$. In fact, the convergence will be cubic, because $g''(x) = 2 \sin(x)$ implies $g''(0) = 0$.
The primary use of this technique is to construct problems which will allow students to apply the definition of stable/unstable fixed points and study the convergence of functional iterations. The need to evaluate the derivative at the fix point significantly reduces the practical utility of this technique. If we cannot evaluate the derivative, then we might as well use the hybrid method obtained by combining bisection with the secant method to achieve ensure convergence which is ultimately superlinear.
