Suppose f(x) is a polynomial with real coefficient and $f(x)>=0$ $x \in R$ let $g(x)=f(x)+f^{'}(x) +f^{"}(x)+..$. prove that $g(x)>=0 for all x\in R$

Question

I am very clueless about how to begin with the problem except the fact that if $f(x)$is positive doesn't imply that derrivative of f(x) is positive .

Answer 1

Let $$f(x)=ax^2+bx+c, a>0, ~~ b^2-4ac\le 0$$ Then $$g(x)=f(x)+f'(x)+f''(x)= ax^2+bx+c+2ax+b+2a= ax^2+(2a+b)x+(2a+b+c)$$ the discriminant of this quadratic is $D=b^2-4ac-4a^2\le0$ fo $g(x)$ is also positive definite for all real values of $x$.

Answer 2

Hints: $g(x)=f(x)+f'(x)+f''(x)+...$ Now differentiate both side the you get

$\Rightarrow g'(x)=f'(x)+f''(x)+f'''(x)+...$

$\Rightarrow g(x)=f(x)+g'(x)$ (I think you can finish it)