Question

f:x maps onto ax+b, f^3:x maps onto 27x+26, find value of a and b.

Answer 1

You can think of the first function as \(f(x)=y=ax+b\). I'm not sure about the notation for the second function... It could mean either \([f(x)]^3\) or \((f\circ f\circ f)(x)=f(f(f(x)))\). My guess would be the latter, because cubing a binomial does not give you another binomial.

If this is the case, then
\[\begin{align*}f(f(f(x)))&=f(f(ax+b))\\&=f(a(ax+b)+b)\\&=a(a(ax+b)+b)+b\\&=a(a^2x+ab+b)+b\\
&=a^3x+a^2b+ab+b
\end{align*}\]
You want to find \(a\) and \(b\) such that .
\[\begin{align*}a^3&=27&\text{matching }x\text{ power terms}\\
a^2b+ab+b&=26&\text{matching constant terms}\end{align*}\]
The first equation gives \(a=3\), so subbing that into the second equation you can directly solve for \(b\):
\[9b+3b+b=13b=26~~\iff~~b=2\]
So, the first function is \(f:x\to(3x+2)\).