$f(x) = ax^3 + bx^2 + cx + d\sin(x)$ . Find condition for which function is one one given that $a>0$.

Question

I approached the question as follows: For the function to be one one it must be strictly increasing or strictly decreasing, hence the derivative must be positive or negative for all $x$ belongs to $\mathbb{R}$. $f'(x) = 3ax^2 + 2bx + c + d\cos(x)$ for strictly increasing, $f'(x)>0$ minimum value of the quadratic is $-D/4a$ which equals to $c-b^2/3a$. Now, (quadratic + $d\cos(x)$) should be positive, hence for limiting case minimum value of quadratic + minimum value of dcosx must be positive $c-b^2/3a -d>0$.

This is the answer given in almost all the books. But I am confused as to why they have not considered the strictly decreasing case. Also, I would like to know that in such questions do we always work by taking $2$ separate cases of strictly increasing and strictly decreasing or there is some logic behind taking only a single case and proceeding?

Answer 1

If $a>0$, it's clear that the function can't be strictly decreasing.

On the other hand, even without that condition, notice there is a bijective correspondence between the set of functions of the given form which are strictly increasing, and the set of functions of the given form which are strictly decreasing, given by $f\mapsto -f$. In the strictly increasing case, the condition $f'>0$ becomes $(-f)'<0$ in the strictly decreasing case, which are algebraically equivalent. Therefore it suffices to look only at the strictly increasing functions.