Can we say that $\forall x\in \Bbb R$, $g(x)>0$, $g(x)<0 $ or that $(\exists x\in \Bbb R) g(x)=0$.

Question

Let $ (a,b,c)\in \Bbb R^3 $ with $ a\ne 0 $,

$$f(x)=ax^2+bx+c$$ and $$g(x)=f(x)+f'(x)+f''(x).$$

We assume that $$(\forall x\in \Bbb R)\;\; f(x)>0$$ What can we conclude $$1. \;\;(\forall x\in \Bbb R)\;\;g(x)>0$$ $$2. \;\;(\forall x\in \Bbb R)\;\; g(x)<0$$ or $$3. \;\; (\exists x\in \Bbb R)\;:\; g(x)=0$$

I used the fact that $$(\forall x\in \Bbb R)\;\; f(x)>0 \implies \delta=b^2-4ac<0$$ then, i wrote $ g(x) $ as $$g(x)=ax^2+(2a+b)x+2a+b+c$$ the discriminant is $$\Delta=(2a+b)^2-4a(2a+b+c)$$ $$=b^2-4ac-4a^2=\delta-4a^2<0$$ i concluded that the sign of $ g(x) $ is constant but i cannot say if it is positive or negative.

Any idea will be appreciated.

Answer 1

We can conclude $g(x)>0$ everywhere. If $f(x)>0$ then $a>0$, and since the coefficient of $x^2$ in $g(x)$ is also $a$, and the discriminant of $g$ is negative, we must have $g(x)>0$ for all $x$.

Answer 2

As J. W. Tanner said in the comments, $f(x)>0,$ so $a>0$, so $g(x) = ax^2+ ...>0$.

Answer 3

Using the "vertex form" of the polynomials, we are taking $$ f(x) \ \ = \ \ ax^2 \ + \ bx \ + \ c \ \ = \ \ a·\left(x \ + \ \frac{b}{2a} \right)^2 \ + \ \left(c \ - \ \frac{b^2}{4a} \right) \ > \ 0 \ \ , $$ which then requires $ \ c - \frac{b^2}{4a} \ > \ 0 \ \ . $ [Naturally, this is equivalent to the discriminant condition.]

For $ \ g(x) \ = \ f(x) + f'(x) + f''(x) \ = \ ax^2 + (2a+b)x + (2a+b+c) \ \ , $ we have $$ g(x) \ \ = \ \ a·\left(x \ + \ \frac{(2a \ + \ b)}{2a} \right)^2 \ + \ \left( \ [2a + b + c] \ - \ \frac{(2a \ + \ b)^2}{4a} \right) $$ $$ = \ \ a·\left(x \ + \ \left[ \ \frac{ b }{2a} + 1 \ \right] \right)^2 \ + \ \left( \ [2a + b + c] \ - \ \left[a + b + \frac{b^2}{4a} \right] \ \right) $$ $$ = \ \ a·\left(x \ + \ \left[ \ \frac{ b }{2a} + 1 \ \right] \right)^2 \ + \ \left( \ a + \left[ \ c - \frac{b^2}{4a} \ \right] \ \right) \ \ . $$ The vertex of the parabola corresponding to $ \ g(x) \ $ is therefore always one unit "to the left" and $ \ a \ $ units "above" the vertex for $ \ f(x) \ $ with $ \ a > 0 \ \ , $ guaranteeing that $ \ g(x) > 0 \ \ . $ [I find that working with quadratic polynomials in this form is frequently simpler than working with discriminants, which can sometimes pose problems in interpretation.]