Investigating monotone and bounded nature of a function. If $$f(x)=x^3+bx^2+cx+d$$ and $0<b^2<c$, then $f(x)$ in $(-\infty,\infty)$


*

*is increasing

*is decreasing

*is bounded

*has real maximum


I solved till $f'(x)=3x^2+2bx+c$ and $f ''(x)=6x+2b$, now what should I do?
 A: Hint
Consider the derivative $f'(x)=3x^2+2bx+c$; solving, for $x$, $f'(x)=0$ (which is a quadratic equation) shows that it cancels at $$x_{\pm}=\frac{1}{3} \left(-b\pm\sqrt{b^2-3 c}\right)$$ But you are told that $0<b^2<c$; so, what about the roots in the real domain ?
I am sure that you can take from here.
A: It is a good idea to complete the square as

$$ f'(x)=3x^2+2bx+c = 3(x+b/3)^2+c-\frac{b^2}{3}. $$

Now see this $(x+b/3)^2 \geq 0$. If you can prove that $ c-\frac{b^2}{3} > 0 $ then you can prove that the function is increasing.
A: We can eliminate 2 and 3 rather easily.  Since the coefficient of $x^3$ is positive, we have $\lim_{x\to-\infty}f(x)=-\infty$ and $\lim_{x\to\infty}f(x)=\infty$, so it's neither bounded nor decreasing.  So we want to know if it's increasing or has a real maximum, both of which suggest looking at the derivative.  To see if it has any roots, we set
$$f'(x)=3x^2+2bx+c=0$$
The discriminant of this quadratic is $4b^2-12c<4c-12c<0$.  Since our discriminant is negative, the derivative has no roots and we have no local minima or maxima.  This rules out the fourth choice and confirms the first.
