The function is $f(x) = x^3 + ax^2 + bx +c$
I am clueless of what to do with the fact the information $a^2 > 3b$ having to do with only having one local maximum and minimum.
What I know is to do the second derivation test but then I would need the a b and c, Any ideas?
$f'(x)=3x^2+2ax+b=0$, this quadratic will have two real roots for max and min in $f(x)$ if $B^2 > 4AC$. So here $4a^2 >12b \implies a^2 >3b$.
Since $f(x)$ is a polynomial, it is unbounded when $x\rightarrow \pm \infty$. Therefore the maximum and minimums (extremums) of the function only occurs at critical points where $f'(x)=0$. We have $f'(x)=3x^2+2ax+b$. This is a quadratic equation, number of roots of $f'(x)$ tells us the number of extremums of $f(x)$. Therefore $\Delta=(2a)^2-4(b)(3)=4a^2-12b=4(a^2-3b)>0$. Therefore we have two extremums, the coordinate of extremums is $x_{1,2}=\frac{-2a\pm \sqrt{4(a^2-3b)}}{2(3)}=-a\pm\sqrt{a^2-3b}$. To find out which one is minimum, we use the second derivative test (at minimums, we have $f''(x)>0$). Since$f''(x)=6x+2a$.
$$f''(x_{1,2})=6\left(\frac{-2a\pm \sqrt{4(a^2-3b)}}{2(3)}\right)+2a=\pm \sqrt{4(a^2-3b)}$$
We see that for sure, we have one minimum and one maximum.
