1. functions prob

if f(x) = ax + b , and f^3(x) = 64x + 21 , find the rule of f^n(x).i appreciate everyone's help.

2. Hello, muslim!

$\displaystyle \text{If }f(x) \,=\, ax + b,\,\text{ and }f^3(x) \,=\, 64x + 21,\:\text{find the rule of }f^n(x).$

$\displaystyle f(x) \:=\:ax+b$

$\displaystyle f^2(x)\;=\;f(f(x)) \;=\;f(ax+b)\;=\;a(ax+b) + b \;=\;a^2x + ab + b$

$\displaystyle f^3(x)\;=\;f(f(f(x))) \;=\;f(a^2x+ab + b) \;=\;a(a^2x+ab+b) + b \;=\;a^3x + (a^2+a+1)b$

Since $\displaystyle f^3(x) \:=\:64x + 21$
. . we have: .$\displaystyle a^3x + (a^2+a+1)b \;=\;64x + 21$

Equating coefficients: .$\displaystyle \begin{array}{cc}a^3 \;=\;64 & [1] \\ (a^2+a+1)b \;=\;21 & [2] \end{array}$

From [1]: .$\displaystyle a^3 \:=\:64\quad\Rightarrow\quad a \,=\,4$

Substitute into [2]: .$\displaystyle (4^2+4+1)b \:=\:21 \quad\Rightarrow\quad 21b \:=\:21 \quad\Rightarrow\quad b \:=\:1$

Hence:. . $\displaystyle f(x) \:=\:4x + 1$

We have:

. . $\displaystyle \begin{array}{ccccccc} f(x) &=& & =& 4x + 1 \\ f^2(x) &=& f(4x+1) &=& 16x + 5 \\ f^3(x) &=& f(16x+5) &=& 64x + 21 \\ f^4(x) &=& f(64x+21) &=& 256x + 85 \end{array}$

Therefore: .$\displaystyle f^n(x) \;=\;4^nx + \frac{4^n-1}{3}$

3. thanx

Originally Posted by Soroban
Hello, muslim!

$\displaystyle f(x) \:=\:ax+b$

$\displaystyle f^2(x)\;=\;f(f(x)) \;=\;f(ax+b)\;=\;a(ax+b) + b \;=\;a^2x + ab + b$

$\displaystyle f^3(x)\;=\;f(f(f(x))) \;=\;f(a^2x+ab + b) \;=\;a(a^2x+ab+b) + b \;=\;a^3x + (a^2+a+1)b$

Since $\displaystyle f^3(x) \:=\:64x + 21$
. . we have: .$\displaystyle a^3x + (a^2+a+1)b \;=\;64x + 21$

Equating coefficients: .$\displaystyle \begin{array}{cc}a^3 \;=\;64 & [1] \\ (a^2+a+1)b \;=\;21 & [2] \end{array}$

From [1]: .$\displaystyle a^3 \:=\:64\quad\Rightarrow\quad a \,=\,4$

Substitute into [2]: .$\displaystyle (4^2+4+1)b \:=\:21 \quad\Rightarrow\quad 21b \:=\:21 \quad\Rightarrow\quad b \:=\:1$

Hence:. . $\displaystyle f(x) \:=\:4x + 1$

We have:

. . $\displaystyle \begin{array}{ccccccc} f(x) &=& & =& 4x + 1 \\ f^2(x) &=& f(4x+1) &=& 16x + 5 \\ f^3(x) &=& f(16x+5) &=& 64x + 21 \\ f^4(x) &=& f(64x+21) &=& 256x + 85 \end{array}$

Therefore: .$\displaystyle f^n(x) \;=\;4^nx + \frac{4^n-1}{3}$

but sorobon , how you identify [(4^n) - 1 / 3]??can you show me the steps??