Answer
$(f o g)(x)$ is the same as $f(g(x))$. However, $(f * g)(x)$ is the same as $f(x)*g(x)$.
Work Step by Step
Let $f(x)=x^2+2$, and let $g(x)=x-3$.
$(f o g)(x) = f(g(x))$
$f(g(x)) = f(x-3)$
$f(x-3) = (x-3)^2+2$
$f(x-3) = (x-3)(x-3) +2$
$f(x-3) = x*x +x*-3 + x*-3 + (-3)(-3) +2$
$f(x-3) = x^2-3x-3x+9+2$
$f(x-3) = x^2-6x+11$
$(f * g)(x) = (x^2+2)(x-3)$
$(f * g)(x) = x^2*x+x^2*-3+2*x+2*-3$
$(f * g)(x) = x^3-3x^2+2x-6$
$x^2-6x+11$ is not the same as $x^3-3x^2+2x-6$, so the two functions are different.
