= Solution
{c}
By the <mean value theorem>, $|f(x)-f(y)|\leq A|x-y|$ for all real $x,y$. Thus $f$ is a <contraction mapping> on the <complete metric space> $\mathbb R$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\leq A^{n-1}|x_2-x_1|$. For $m>n$, the <triangle inequality> and a <geometric series> yield
$$
|x_m-x_n|\leq\frac{A^{n-1}}{1-A}|x_2-x_1|\longrightarrow0.
$$
Hence $(x_n)$ is a <Cauchy sequence> and <completeness of the real numbers> supplies a <limit> $x$. The <Lipschitz continuity> of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a <fixed point>, independently of the initial value. If $y$ were another, $|x-y|\leq A|x-y|$ would force $x=y$. This proves the <contraction mapping theorem> in this setting, with error estimate
$$
\boxed{|x-x_n|\leq\frac{A^{n-1}}{1-A}|x_2-x_1|.}
$$
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