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Let $f(x)$ be a quadratic expression which is positive for all real values of $x$. If $g(x)=f(x)+f'(x)+f''(x)$, then for any real $x$, $$(A)\, g(x)<0,\quad (B)\,g(x)>0,\quad (C)\, g(x)=0,\quad (D)\;g(x)\ge 0.$$

My work so far. Let $f(x)=ax^2+bx+c$ be the quadratic polynomial. Then $b^2-4ac<0$ and $$g(x)=ax^2+x(b+2a)+(c+b+2a).$$ Now I am stuck. I am not able to pick the right answer.

Robert Z
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Brahmagupta
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  • Have you tried an example? – lulu Jun 19 '17 at 15:09
  • Related: https://math.stackexchange.com/questions/789848/prove-that-gx-0 – Winther Jun 19 '17 at 15:14
  • The first term of $f(x) + f'(x) + f''(x)$ is positive by hypothesis, and the third term must be positive since the parabola must open upward. The second term must be negative for some values of $x$ and positive for others, given the way parabolas look. So if one of the answers is correct, it can only be (B) or (D). – Michael Hardy Jun 19 '17 at 15:14
  • You've done most of what needs to be done and Robert Z's answer does the rest. – Michael Hardy Jun 19 '17 at 15:20

2 Answers2

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$f(x)=ax^2+bx+c$. $f>0 \Rightarrow a>0$.

$f(x)+f^\prime(x)+f^{\prime\prime}(x)=ax^2+(2a+b)x+c+b+2a=g(x)$

Reformulating $g$ in terms of $x+1$ gives $$a(x+1)^2+ (2a+b)(x+1)+c+b+2a-(2ax+a)-(2a+b) = $$ $$a(x+1)^2+ (2a+b)(x+1)+c-2a(x+1)+a = $$ $$a(x+1)^2+b(x+1)+c+a=g(x)$$.

So $g(x) = f(x+1)+a$, so $g(x)$ is $f(x)$ translated by 1 to the left and by $a$ upwards.

$f>0 \Longrightarrow g>0$

Χpẘ
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Hint. Note that the discriminant of the quadratic polynomial $g$ is $$(b+2a)^2-4a(c+b+2a)=(b^2-4ac)-4a^2<0.$$ Moreover the coefficient of $x^2$ of $g$ is $a>0$. What is the sign of $g$?

Robert Z
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