Question

If $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$ then find range of $f(x)$.

Answer 1

SOLUTION : $1+\sqrt{x}=t \quad \Rightarrow \quad x=(t-1)^2 \\$

$f(t)=3+2(t-1)+(t-1)^2=t^2+2 \\$

$f(x)=x^2+2 \\$

$\text { Range }=(2, \infty)$