Exercise 5.22

(a) Suppose $f(a)=a$ and $f(b)=b$ for $a<b$. By Theorem 5.10, there is a point $t$, $a<t<b$, such that $f^{\prime }(t)=\left(f(b)-f(a)\right)∕(b-a)=1$, contradicting $f^{\prime }(t)\ne 1$ for all real $t$.

(b) If $t=f(t)=t+{(1+e^t)}^{-1}$, then ${(1+e^t)}^{-1}=0$, which is impossible. We have

Since $f^{″}(t)$ has a single zero, at $t=0$, $f^{\prime }(t)$ decreases from $\underset{t\to -\infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}^{\prime }(t)=1$ to $f^{\prime }(0)=3∕4$, then increases to $\underset{t\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}^{\prime }(t)=1$, so that the range of $f^{\prime }$ is $[3∕4,1)$.

(c) Since $x_{n+1}=f(x_n)$ and $x_n=f(x_{n-1})$, by Theorem 5.10 there is a point $t$ between $x_{n-1}$ and $x_n$ such that

Hence

So for $m$ and $n$ greater than $N$, we have

Since $A<1$, this last term goes to 0 as $N\to \infty$, so $\{x_n\}$ is a Cauchy sequence, converging to $x$. Since $f$ is differentiable, it is continuous, hence

that is, $x$ is a fixed point of $f$.

(d) This zig-zag path is described in $R^2$ with respect to the graph of $f$ in that space. It starts with the point $\left(x_1,x_2=f(x_1)\right)$ on the graph of $f$, goes horizonatally until it meets the diagonal $y=x$ at $(x_2,x_2)$ then goes vertically until it hits the graph of $f$ again at $\left(x_2,x_3=f(x_2)\right)$, and so forth. It will zig-zag or spiral to the point on the graph of $f$ corresponding to a fixed point of $f$, where it crosses the diagonal $y=x$.