Suppose that $g(x)=1+ \sqrt{x}$ and $f(g(x)) = 3 +2\sqrt{x}+x$. Then find the function f(x).

$3+x^2$ $x^2+2x$ $2+x^2$ $x^2+3x$

$2+x^2$

$g(x)=1+ \sqrt{x}$ and $f(g(x)) = 3 +2\sqrt{x}+x$ $∴f(1+\sqrt{x})=3+2\sqrt{x}+x$ Put $1+ \sqrt{x}=y$ or $x=(y-1)^2$. Then, $f(y)=3+2(y-1)+(y-1)^2 = 2 + y^2$ $∴f(x)=2+x^2$