= Solution
{c}
Set $H(x)=g(x)-x$. This <function> is <continuous> on $[a,b]$. Since $g$ takes values in $[a,b]$,
$$
H(a)=g(a)-a\ge0,\qquad H(b)=g(b)-b\le0.
$$
The <intermediate value theorem> therefore supplies $\xi\in[a,b]$ with $H(\xi)=0$, equivalently $g(\xi)=\xi$. This includes the possibility that one endpoint is already a <fixed point>.
For uniqueness, the <mean value theorem> gives, for any $x,y\in[a,b]$,
$$
|g(x)-g(y)|\le\gamma|x-y|.
$$
If both $\xi$ and $\eta$ are <fixed points>, then
$$
|\xi-\eta|=|g(\xi)-g(\eta)|\le\gamma|\xi-\eta|.
$$
Since $\gamma<1$, this forces $|\xi-\eta|=0$. Thus \b[there exists exactly one <fixed point>]. When $a=b$, the conclusion follows immediately from $g(a)=a$; otherwise the argument above applies. The <contraction mapping> estimate also proves convergence of the associated <fixed-point iteration>, as used next.
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