<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] Not a great deal faster, but you could also use the fact that the local extrema in the cubic function are symmetric about its inflection point, given by $ \ f''(x) \ = \ 6x +  2a \ = \ 0 $ $ \Rightarrow \ x_{infl} \ = \ -\frac{a}{3} \ \ . $  Since the local maximum is on the $ \ y-$axis (if the diagram is to be at all trusted), then the local minimum is at $ \ x_{min} \ = \ -\frac{2a}{3} \ \ . $  The function value at the inflection point is $ \ f\left(-\frac{a}{3} \right) \ = \ -\frac{a^3}{27} + \frac{a^3}{9} - \frac{ab}{3} + 4 \ = \ 2  \ \ . $  You will want to use your observation that $ \ f'(0) \ = \ 0 \ \Rightarrow \ b \ = \ 0 \ \ , $ so we have $ \ \frac{2·a^3}{27} \ = \ -2 \ \Rightarrow \ a^3 \ = \ -27 $ $ \Rightarrow \  x_{min} \ = \ -\frac{2·[-3]}{3} \ = \ +2 \ \ . $
Your educated guess that $ \ x^3 - 3x^2 + 4 \ = \ (x + 1)·(x - 2)^2 \ $ is correct; acting on that during the exam depends on how much of a gambler you are:  Viete tells you that $ \ c \ = \ 4 \ = \ -r·s^2 \ \ , $ so $ \ r \ = \ -1 \ \ , \ \ s \ = \ +2 \ $ would be worth the risk if you're desperate for time...
ADDENDUM:  It might also be noted that once you see that $ \ c \ = \ 4 \ \ , $ the Rational Zeroes Theorem lets you "reject" choices (3) and (4) (now... does $ \ \frac{1}{2} \ $ seems plausible...?).
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