## Calculus: Early Transcendentals (2nd Edition)

The solution is $$f(x)=3x-2.$$
On the right side we have a polynomial of the second degree. On the left we have $(f(x))^2$. Since this has to be a polynomial of 2nd degree we have to demand that $f$ is a first degree polynomial function i.e. it is of the form of $f(x)=ax+b$. Now lets determine $a$ and $b$. $$(f(x))^2=(ax+b)^2=(ax)^2+2abx+b^2=a^2x^2+2abx+b^2=9x^2-12x+4.$$ Equating the coefficinets multiplyng same powers of $x$ we get $$a^2=9\Rightarrow a=3;$$ $$2ab=-12\Rightarrow2\cdot3b=-12\Rightarrow b=-2;$$ The last equation $$b^2=4$$ is now used just for checking and indeed $2^2=4$. Finally, $$f(x)=3x-2.$$