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

[tex]\bf \begin{array}{llll} term&&value\\ \text{\textemdash\textemdash\textemdash}&&\text{\textemdash\textemdash\textemdash}\\ 0&&1\\ 1&&3(1)+5\implies 8\\ 2&&3(8)+5\implies 29\\ 3&&3(29)+5\implies \boxed{?} \end{array}[/tex]

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ThonyBossa ThonyBossa · Answer 2 · 2019-08-17T03:52:39+02:00

Answer:

The value for the third iterate is 92.

Step-by-step explanation:

We have the next :

[tex]f(x)=3x+5[/tex] ; [tex]x_{0}=1[/tex]

To solve it, let's go to replace the initial value on the equation and the result will be the first iterate. We will repeat two times more to find the third iterate, that is:

[tex]x_{1}[/tex] ; First iterate

[tex]f(x_{0})=3*(1)+5[/tex]

[tex]f(x)=8=x_{1}[/tex]

[tex]x_{2}[/tex] ; Second iterate

[tex]f(x_{1})=3*(8)+5[/tex]

[tex]f(x)=29 = x_{2}[/tex]

[tex]x_{3}[/tex] ; Third iterate

[tex]f(x_{2})=3*(29)+5[/tex]

[tex]f(x)=92 =x_{3}[/tex]