If $f$ is a quadratic and $f(x)>0\;\forall x$, and $g= f + f' + f''$, prove $g(x)>0\; \forall x$ If $f(x)$ is a quadratic expression such that $f(x)>0\;\forall x\in\mathbb{R},$ and if $g(x)=f(x)+f'(x)+f''(x),$
Then prove that $g(x)>0\; \forall \; x\in \mathbb{R}$.
$\bf{My\; Trial \; Solution::}$ If $f(x)>0\;\forall x\in \mathbb{R}$. Then function $f(x)$ has Minimum value .
Let Minimum occur at $x=x_{0}$. Then $f(x)_{Min} = f(x_{0})>0$
Now Given $g(x) = f(x)+f'(x)+f''(x)$. Then $g'(x) = f'(x)+f''(x)+f'''(x).$
Now at $x=x_{0}\;\;,$ Value of $g(x_{0}) = f'(x_{0})+f''(x_{0})+f'''(x_{0}) = 0+f''(x_{0})+f'''(x_{0})$
Now I did not understand How can i solve after that, 
Help me 
Thanks
 A: Suppose $f(x)=x^2+ax+b$ with $b=f(0)>0$ and $a^2<4b$. Then,
$$
g(x)=x^2+(a+2)x+(2+a+b).
$$
We note that $g(0)=2+a+b=1+f(1)>0$ and
$$
(a+2)^2-4(2+a+b)=a^2+4a+4-8-4a-4b=(a^2-4b)-4<0
$$
so $g$ is never $0$ for real $x$. You now can infer that $g$ is always positive.
The more general case $f(x)=C(x^2+ax+b)$ is the same. It just adds the constant $C>0$ to everything.
A: Since $f$ is positive, we can complete the square and write $f(x)=a(x-h)^2+k$ with $a,k > 0$. Then 
\begin{eqnarray}
g(x)&=&a(x-h)^2+k+2a(x-h)+2a\\
&=&a(x-h)(x-h+2)+k+2a\\
&=&a[(x-h+1)^2-1]+k+2a\\
&=&a(x-h+1)^2+k+a
\end{eqnarray}
is a similarly positive quadratic expression.
ETA: A goofy alternative way of phrasing this would be to say:


*

*Since $f$ is quadratic, all its higher-order derivatives vanish. 

*So, if $D$ is the differential operator, then $f(x)+f'(x)+\frac{f''(x)}{2}=(e^Df)(x)=f(x+1)$.

*So $g(x)=f(x+1)+\frac{f''(x)}{2}$. But if $f$ is quadratic and everywhere positive, then its second derivative must be positive; thus $g$ is everywhere positive.

A: Without loss of generality, we can assume that $f(x)= (x-a)^2 + b^2$. 
Note that $g$ is also a quadratic polynomial. We find the critical point of $g$. 
$$
g'(x) = f'(x) +  f''(x)$$
Note also that third derivative of $f$ is zero.
Then we see that the critical point is $x=a-1$, and also it gives the minimum value of $g$. 
Plug in the value to $g$, then we have
$$
g(a-1) = f(a-1) + g'(a-1) = f( a-1) > 0.$$
Therefore $g(x)>0$ for all $x\in \mathbb{R}$. 
