Question

Find the third iterate, x3, of the function f(x) = 3x + 5 for an initial value of x0 = 1.

Answer 1

I assume by "iterate" you mean \(x_n=f(x_{n-1})\). Under this assumption, you have
\[x_1=f(x_0)=3x_0+5=3\times1+5=8\]followed by\[x_2=f(x_1)=16\]Try working out \(x_3\) on your own.

Answer 2

it is not right

Answer 3

why xn = f(x n-1)

Answer 4

it is 92
For x0
f(1)= 8
For x1
f(8) = 29
For x2
f(29) = 92
for x3
f(92)= 281

so the input x3 of the function is what we want

Answer 5

> "why xn = f(x n-1)"

That's why I said "assuming"; I wasn't sure myself, but that seems like a reasonable guess for what "iterate" means in this context. Given some recursive function \(x_n=f(x_{n-1},x_{n-2},\ldots,x_1,x_0)\), the \(k\)th iterate could mean the \(k\)th term in the sequence \(\{x_0,x_1,\ldots,x_{k}\}\).

In this case,
\[\begin{array}{c|c|cl}
n&f(x_{n-1})&x_n&(n\text{th iterate})\\
\hline
0&\text{N/A}&\color{red}1&0\text{th iterate}\\
1&f(1)=3(\color{red}1)+5&\color{blue}8&1\text{st iterate}\\
2&f(8)=3(\color{blue}8)+5&\color{green}29&2\text{nd iterate}\\
3&f(29)=3(\color{green}29)+5&\fbox{92}&3\text{rd iterate}
\end{array}\]