Question 1182908
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Given that f : x maps to ax + b and f^3 : x maps to 27x + 26, 
Find the value of a and of b, Find an expression for f^4
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<pre>
f^2(x) = a*(ax+b) + b = (a^2)x + ab + b.


f^3(x) = (a^2)*(ax+b) + ab + b = (a^3)x + (a^2)*b + ab + b = 27x + 26.


From the last equation,  a^3 = 27;  hence,  a = {{{root(3,27)}}} = 3.


Then we have


    3^2*b + 3b + b = 26,   or   9b + 3b + b = 26,   13b = 26,  b = 26/13 = 2.



So,  f(x) = 3x + 2.


Finally,  f^4(x) = 27(3x+2) + 26 = 81x + 54 + 26 = 81x + 80.    <<<---===  I just edited this line after the notice by @robertb.   Thanks (!)


<U>ANSWER</U>.  a= 3;  b= 2;   f^4(x) = 81x + 80.
</pre>

Solved.