Chapter 1, Problem 20P

(a)

To determine

To find: The expression for $f_n(x)$.

Answer to Problem 20P

Solution:

The expression is, $f_n(x)=\frac{n+1-nx}{n+2-(n+1)x}$.

Explanation of Solution

Given:

$f_0(x)=\frac{1}{(2-x)}$.

$f_{n+1}-f_0(f_n(x))$  for $n=0,1,\mathrm{2...}$

Calculation:

Base case: $n=1$.

To prove that the statement is true for $n=1$.

$\begin{array}{l}{f}_{n+1}=f_0(f{(x)}_1)\\ f_2=f_0(f{(x)}_1)\end{array}$

Therefore, the statement is true for $n=1$.

Induction hypothesis: $n=k$

Assuming that the claim is true for $n=k$.

That is, $f_{k+1}(x)=f_0(f_k(x))$

Inductive step: $n=k+1$

To prove the statement is true for $n=k+1$.

$f_0(x)=\frac{1}{(2-x)}$

Substitute $n=1$ ,

$f_1(x)=f_0\left(\frac{1}{(2-x)}\right)$

$\begin{array}{l}{f}_1(x)=\frac{1}{2-\left(\frac{1}{(2-x)}\right)}\\ f_1(x)=\frac{2-x}{3-2x}\end{array}$

Substitute $n=2$ ,

$\begin{array}{l}{f}_2(x)=\frac{1}{2-\left(\frac{2-x}{3-2x}\right)}\\ f_2(x)=\frac{3-2x}{6-4x-(2-x)}\\ f_2(x)=\frac{3-2x}{4-3x}\end{array}$

Substitute $n=3$ ,

$\begin{array}{l}{f}_3(x)=\frac{1}{2-\left(\frac{3-2x}{4-3x}\right)}\\ f_3(x)=\frac{4-3x}{8-6x-(3-2x)}\\ f_3(x)=\frac{4-3x}{5-4x}\end{array}$

Thus, the general formula is $f_n(x)=\frac{n+1-nx}{n+2-(n+1)x}$

The claim is true for $n=k$.

$f_k(x)=\frac{k+1-kx}{k+2-(k+1)x}$

To prove the statement is true for $n=k+1$.

$\begin{array}{l}{f}_{k+1}(x)=f_0(f_k(x))\\ f_{k+1}(x)=f_0\left(\frac{k+1-kx}{k+2-(k+1)x}\right)\\ f_{k+1}(x)=\frac{1}{2-\frac{k+1-kx}{k+2-(k+1)x}}\\ f_{k+1}(x)=\frac{k+2-(k+1)x}{k+4-2\left(\left(k+1\right)x\right)-k+1-kx}\end{array}$

$f_{k+1}(x)=\frac{k+2-\left(k+1\right)x}{k+3-\left(k+2\right)x}$

Therefore, that is true for$k+1$ .

Thus, $f_n\left(x\right)$  is true for all positive integer $n$.

(b)

To determine

To graph: The function$f_0,f_1,f_2,f_3$ .

Explanation of Solution

Graph:

The graph of the functions are shown below in Figure 1.

From Figure 1, if the value of n increased, then the vertical asymptotes moves to $y=1$.

If the value of $n$  increased, then the horizontal asymptotes moves to $x=1$.

The $x$ - intercept for $f_{n+1}$  is the value of vertical asymptote$f_n$

The $y$ - intercept for $f_n$  is the value of vertical asymptote $f_{n+1}$.