Suppose that $g(x)=1+ \sqrt{x}$ and $f(g(x)) = 3 +2\sqrt{x}+x$. Then find the function f(x).
Answer & explanation
Correct answer: option 3
$g(x)=1+ \sqrt{x}$ 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,
$f(y)=3+2(y-1)+(y-1)^2 = 2 + y^2$
$∴f(x)=2+x^2$