Prove that a 2 degree polynomial and it's derivative > 0 .

Question

The question is

Let $f(x)$ be a polynomial of degree $2$ and $f(x) > 0$ for all real values of $x$. Prove that $f(x) + f'(x) + f''(x) > 0$ for any value of $x$.

I tried making that kind of a function and started putting values. It's coming right, but I wanted a more rigorous proof

Answer 1

Since $\deg{f(x)}=2$ and $f(x)>0$ for all $x$, let $f(x)=ax^2+bx+c$, with $a>0$ and $b^2-4ac<0$. (Explanation below)

Compute $f'(x), f''(x)$ then,

$$f(x)+f'(x)+f''(x) = ax^2+(2a+b)x+(2a+b+c)$$

Use the discriminant on this quadratic equation.

$$D=(2a+b)^2-4a(2a+b+c)=b^2-4ac-4a^2 < 0$$

Therefore for all $x\in\mathbb{R}$, $$f(x)+f'(x)+f''(x) = ax^2+(2a+b)x+(2a+b+c) > 0$$

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Given a quadratic equation $ax^2+bx+c=0$,

• if $D=b^2-4ac\geq0$, the equation has real roots.

• if $D =b^2-4ac<0$, the equation has no real roots, so there doesn't exist any real values of $x$ that will satisfy $ax^2+bx+c=0$. Therefore, $ax^2+bx+c$ should be either greater than $0$ or less than $0$ for all real values of $x$.

In the second case, for $ax^2+bx+c>0$ to hold, it must be that $a>0$, since if $a<0$, as $x \rightarrow \infty$, $ax^2+bx+c \rightarrow -\infty$

Answer 2

A polynomial of degree $2$ describes a parabola.
$f(x)=a(x-b)^2+c$.

As $f>0$, we need $a,c>0$ (can easily be seen when plotting the graphs).

Thus we get $$f(x)+f'(x)+f''(x)=a(x-b)^2+c+2a(x-b)+2a\\ =a[ (x-b)+1]^2 -a+2a+c >0,$$ as $a,c, [x-b)+1]^2>0$.