Hey guys,
"Find a function g(x) such that fog(x) is a quadratic function"
f(x) = (1+x)^1/2 - 2
(which would be square root (1+x)) - 2
I think this would work.
$\displaystyle f(x)=\sqrt{1+x}-2$
Let: $\displaystyle g(x)=x^4-1$
$\displaystyle (f\circ g)(x)=\sqrt{1+(x^4-1)}-2$
$\displaystyle =\sqrt{x^4}-2$
$\displaystyle =x^2-2$
This is a second degree polynomial, so I guess the function your looking for is $\displaystyle g(x)=x^4-1$