<s>[INST] Given graph of the function $f(x)=x^3+ax^2+bx+c$ What is the $x$ value at the local minimum point? This is a problem from a timed exam, so I prefer approaches that lead to answer quickly.

Graph of the function $f(x)=x^3+ax^2+bx+c$ is as follow. What is the
$x$ value at the local minimum point?
$1)\frac12\qquad\qquad2)2\qquad\qquad3)\frac32\qquad\qquad4)3$


Here is my approach,
Suppose we have minima at $x_0$. We have,$$f(0)=4\Rightarrow c=4$$
$$f'(x)=3x^2+2ax+b\qquad\text{Since $f'(0)=0\rightarrow b=0$}$$Hence $f'(x)=x(3x+2a)$, So $x_0=-\frac{2a}3$. We have $f(x)=x^3+ax^2+4$. $$f(-\frac{2a}3)=0\Rightarrow -\frac{8a^3}{27}+\frac{4a^3}9+4=0\Rightarrow a=-3$$
Finally $x_0=\frac{-2a}3=2$.

Although one can get the answer with this approach in one or two minutes, I'm looking for quicker ways to solve the problem. Actually, first time I tried to solve the problem, I noticed that we can write $f(x)=(x+m)(x-n)^2$ where $m,n\in \mathbb{R}^+$. According to graph of function and noting that $f(0)=mn^2=4$, I can guess it the function is $f(x)=(x+1)(x-2)^2$. But assuming I'm in the exam, at this point should I quickly mark the answer and go to the next question or this approach is unreliable and I got the correct answer with a bit of luck?!
 [/INST] Suppose that a cubic function has critical points at $x = r$ and $x = s$.  IOW, its derivative is zero at these points.  Then
$$f'(x) = k(x - r)(x - s)$$
$$f'(x) = k(x^2 - (r+s)x + rs)$$
$$f(x) = k(\frac{1}{3}x^3 - \frac{1}{2}(r + s)x^2 + rsx) + c$$
But you're given that the $x^3$ coefficient is 1, so $k = 3$.
$$f(x) = x^3 - \frac{3}{2}(r + s)x^2 + 3rsx + c$$
You're also given $r = 0$ and $c = 4$ at the other local extremum, so:
$$f(x) = x^3 - \frac{3}{2}sx^2 + 4$$
But at this local minimum point, $x = s$ and $f(x) = 0$, so:
$$x^3 - \frac{3}{2}x^3 + 4 = 0$$
$$-\frac{1}{2}x^3 = -4$$
$$x^3 = 8$$
$$x = 2$$
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