$f(x) = ax^3 + bx^2 + cx + d,$ with $a > 0. $ If $f$ is strictly increasing, then the function $g(x) = f′ (x) −f′′(x) + f′′′(x)$ is 
QUESTION: 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 the function $g(x) = f′ (x) −f′′(x) + f′′′(x)$ is
$(i)$ zero for some real $x$.
$(ii)$ positive for all real $x$.
$(iii)$ negative for all real $x$.
$(iv)$ strictly increasing.


MY ANSWER: This is an easy question. I just want to know where I went wrong. This is what I did-
$$f'(x)=3ax^2+2bx+c$$$$f''(x)=6ax+2b$$ and, $$f'''(x)=6a$$ therefore, $$g(x)=3ax^2+2bx+c-6ax-2b+6a$$ which is nothing but, $$g(x)=3ax^2+(2b-6a)x+(6a+c-2b)$$ therefore, it represents an upward opening parabola $(\because a>0)$. So we can see that the function is not strictly increasing and not always negative. Now whether or not the function attains zero, depends on the discriminant of the quadratic equation. But it comes out to be $$\sqrt{(2b-6a)^2-4(6a+c-2b)3a}$$ and this becomes too complicated.. also we have another piece of information that $f'(x)=3ax^2+2bx+c>0$. But I couldn't use it anywhere..
I am quite sure there are smarter ways to tackle this one.
Any help will be much appreciated. Thank you so much.
 A: Since it is a multiple-choice question, you have the luxury to plug in values and arrive at the answer by process of elimination. For the easiest case, let $a=1,b=c=d=0.$ So $$g(x)=3x^2-6x+6$$ which is never $0.$ Thus $(ii)$ is your answer.
A: Since $f'(x) = 3ax^2 + 2bx + c \geq 0$, the discriminant of $f'$ should be negative:
$$4b^2 - 4\cdot 3a \cdot c = 4b^2 - 12ac \leq  0$$
Now the discriminant of $g$ equals 
$$(2b-6a)^2 - 4\cdot 3a \cdot (6a+c-2b) = 4b^2 - 24ab + 36a^2 - 72a^2 - 12ac + 24ab = 4b^2 - 12ac - 36a^2$$
but since $4b^2 - 12ac \leq 0$ and $-36a^2 < 0$, this discriminant is striclty negative. 
This implies that $g(x) > 0$ for all $x$, since $g$ can't have any zeroes.
A: We know that $g$ is an U-shaped parabola.  To calculate its vertex, solve 
$$g'(x_0)=0\iff f''(x_0)=f'''(x_0)$$
remembering that $f^{(4)}=0$. The second coordinate of the vertex is
$$g(x_0)=f'(x_0)-f''(x_0)+f'''(x_0)=f'(x_0)\geq0$$
as $f$ is strictly increasing.  
If $f'(x_0)=0$ then $f$ has at $x_0$ a saddle point, that is, $f''(x_0)=0$, too, and that means $0=f''(x_0)=f'''(x_0)=6a>0$.
A: Since $f$ is strictly increasing, the reduced discriminant of $f^\prime$, i.e.:
$\frac{\Delta^\prime}{4} = b^2 - 3ac\;, $
satisfies $\frac{\Delta^\prime}{4} \leq 0$.
Now, the reduced discriminant of $g$, i.e.:
$\frac{\Delta}{4} = (b-3a)^2 - 3a(6a + c - 2b) = b^2 - 6ab + a^2 - 18a^2 - 3ac + 6ab = \frac{\Delta^\prime}{4} - 17a^2\;,$
satisfies $\frac{\Delta}{4} < 0$.
Therefore, $g$ has to have the same sign of $a$.
