Consider that $f:\mathbb R \to \mathbb R$
i) Let $$f(x)=x^3 + x^2 + ax + 4$$ be bijective find $a$
ii) let$$f(x) = ax^3+bx^2+cx+d$$ be bijective, then find the condition
For the first part, the book gives the solution as the inequality that $f'(x)\ge0\implies D\le0$ where $D$ is the discriminant of $f'(x)$. But I don't get why? If$D\le0$ then $f'(x)$ doesn't exist
The solution is save for the second part too. What???!!!
On
For the first one $f'(x)= 3x^2+2x+a$. If the discrimant $D\le 0$ then it means that it has no solution or at least one. Meaning that it has the same sign all over it's set of definition, therefore it is monotonic. So you set:
$$D\le 0 \Rightarrow 4-12a \le 0$$
The same happens with the second one.
If the discriminant of $f'(x)=3x^2+2x+a$ is $<0$, then the derivative cannot change sign, hence $f$ is strictly monotonic. If the discriminant $=0$, then $f'$ vanishes only at one point and is positive everywhere else, so it is strictly monotonic as well.
Since $f$ has limit $-\infty$ at $-\infty$ and $\infty$ at $\infty$, we can conclude that $4-12a\le0$ (that is, $a\ge 1/3$) implies the function $f$ is bijective.
On the other hand, if $a<1/3$, the derivative $f'(x)$ vanishes twice, being positive up to the least root, negative between the roots and positive past the greatest root. Hence it is not bijective, having both a local maximum and a local minimum.
The fact that the discriminant of a degree two polynomial is negative just means the polynomial has no roots. In this case, no root means the derivative (which is continuous) cannot take positive and negative values, so it is either everywhere positive or everywhere negative.