Let f(x) be a polynomial function of degree 2 such that f(x) > 0 for all $x \in R$. If g(x) = f(x) + f'(x) + f''(x) for all x, then

$g(x)>0$ for all x

Let $f(x)=a x^2+b x+c$. Then, f(x) > 0 for all $x \in R$

$\Rightarrow a>0$  and  $b^2-4 a c<0$

Now,

$g(x)=f(x)+f'(x)+f''(x)$ for all x

$\Rightarrow g(x)=a x^2+b x+c+2 a x+b+2 a$

$\Rightarrow g(x)=a x^2+x(2 a+b)+2 a+b+c$

Let D be the discriminant of g(x). Then,

$D =(2 a+b)^2-4 a(2 a+b+c)$

$\Rightarrow D =-4 a^2+b^2-4 a c$

$\Rightarrow D =-4 a^2+\left(b^2-4 a c\right)<0$              [∵ b² - 4ac < 0]

Thus, g(x) is a quadratic polynomial such that coefficient of $x^2>0$ and, D < 0. Therefore, g(x) > 0 for all $x \in R$.