Prove that $g(x) > 0$

Question

If $f(x)$ is a quadratic expression such that $f(x) > 0,\ x\in\mathbb{R}$ and if $g(x)= f(x) + f'(x) + f''(x)$, then prove that $g(x) > 0, \ x\in\mathbb{R}$.

Answer 1

Let $f(x) = ax^2 + bx + c$, then $f'(x) = 2ax + b$, and $f''(x) = 2a$. So:

$g(x) = ax^2 + (2a + b)x + 2a + b + c$.

Let $x = -\dfrac{c}{b}$, then $f\left(-\dfrac{c}{b}\right) = a\cdot \dfrac{c^2}{b^2} > 0$ implies that $a > 0$.

So: $\triangle_g = (2a + b)^2 - 4a(2a + b + c) = -4a^2 + (b^2 - 4ac) < 0$, since

$b^2 - 4ac < 0$ because $f(x) > 0$ $\forall x$. Thus: $g(x) > 0$

Answer 2

A simpler solution:

Take $f(x) = a(x-b)(x-c)$ (note that $a>0$ as $f>0$) and notice that

$g(x) = f(x) + f'(x) + f''(x) = a(x+1-b)(x+1-c) + a = f(x+1) + a > 0$