Now $g(x)=3ax^2+x(2b-6a)+(c-2b+6a)$
The point is as $f(x)$ is strictly increasing, so does $f^{''}(x)$ and so does $f^{'''}(x)$ . Basically $f^{''}(x) \implies$ the rate of increase of $f^{'}(x)$ and $f^{'''}(x)\implies$ the increasing rate of $f^{''}(x)$. Thus when $f^{'}(x)$ goes on strictly increasing, $f^{''}(x)$ increases at a lower rate and $f^{'''}(x)$ increases at a still lower rate.
Thus $f^{'}(x)-f^{''}(x)+f^{'''}(x)>0$ as $f^{'}(x)+f^{'''}(x)$ always adds upto something $> f^{''}(x)$ Why?
$f^{'}(x)>f^{''}(x)\implies f^{'}(x)-f^{''}(x)>0\implies f^{'}(x)-f^{''}(x)+f^{'''}(x)>0$
Thus $g(x)>0$ for all $x \in \mathbb{R}$
Now my question is why is $g(x)$ not strictly increasing?
This question has 4 choices:
A. zero for some $x \in \mathbb{R}$
B. $g(x)>0$ for all $x \in \mathbb{R}$
C. negative for all $x \in \mathbb{R}$
D. strictly increasing.
Now I am confused between B and D. The answer however is B. $g(x)>0$ for all $x \in \mathbb{R}$
But why not D. strictly increasing.?
Please give proper reasoning to answer the question.
Answer to first question: Because it is quadratic function.
It has extremmum. If $a>0$ it has minimum and if $a<0$ it has maximum.
Since $f$ is increasing we have $f'(x)>0$ for all $x$ so $$3ax^2+2bx+c>0$$ for all $x$ so it discriminat $D_{f'} = 4b^2-12ac$ is negative and $a>0$, so discirimnant for $g$ is
$$D_g = (2b-6a)^2 - 12a(c-2b+6a) = 4(b^2-4ac)<0$$
so $g(x)>0$ for all $x$.