Indian Statistical Institute B.Math & B.Stat : Quadratic Equations

Consider the function $f(x) = ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are real numbers with $a > 0$. If $f$ is strictly increasing, then show that the function $g(x) =f′(x) − f′′(x) + f′′′(x)$ is positive for all $x \in \mathbb{R}$. First we calculate the derivatives up to the third order. $f'(x) = 3ax^2+2bx+c, \quad f''(x) = 6ax+2b \quad and \quad f'''(x) = 6a$. It is given that $f$ is strictly increasing which implies $f' > 0$ which in turn implies $3ax^2+2bx+c > 0$. Let $y = 3ax^2+2bx+c$ It is easy to see that $y = 3a \big( x + \frac{b}{3a} \big)^2 + \frac{3ac-b^2}{3a}$. Since $y > 0$ and $a$ is given to be positive $3ac$ must be strictly greater than $b^2$. Note $\big( x + \frac{b}{3a} \big)^2$ is always non-negative. Now $g(x) = f′(x) − f′′(x) + f′′′(x) = 3ax^2+2bx+c -(6ax+2b) + 6a = 3ax^2 + 2x(b-3a)+(c-2b+6a)$ $= 3a\big( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2}+ \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2} \big)$ = $3a\bigg( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2} \bigg) + 3a \bigg( \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2}\bigg)$ $= 3a \big( x + \frac{b-3a}{3a} \big)^2 + \frac{9a^2+3ac-b^2}{3a}$ $3a \big( x + \frac{b-3a}{3a} \big)^2 \geq 0$ for all $x \in \mathbb{R}$. (since $a$ is given to be positive) We have already shown that $3ac > b^2$ therefore $\frac{9a^2+3ac-b^2}{3a} > 0$. Thus $g(x) > 0$ for all $x \in \mathbb{R}$.