Finding function from two point of the curve and two value of the maximum How can address this problem ? The teacher gave us that $ f(x) = ax^3+bx^2+cx+d$ and asked us to find the value of $a$,$b$,$c$ and $d$ knowing that:


*

*$f$ have two extremums $(1)$ and $(-1)$

*$f(0) = 2$ and $f(-2) = 0$


I have no clue how to solve this problem 
 A: Point (1) means the derivative vanishes at $\;x=-1,\,1\;$ :
$$f'(x)=3ax^2+2bx+c\implies\begin{cases}f'(1)=3a+2b+c=0\\{}\\f'(-1)=3a-2b+c=0\end{cases}$$
Point (2) means exactly what it says:
$$\begin{cases}2=f(0)=c\\{}\\0=f(-2)=-8a+4b-2c+d=0\end{cases}$$
Now find out all the coefficietns from the above linear equations and that's all, though it is perhaps a little worrying you have no clue what to do as this subject requires some serious background.
A: At the extrema, it needs to be true that the first order condition holds, i.e.
$$
f'(x) = 3ax^2 + 2bx + c = 0.
$$
You would need to plug in the values for the extrema to get two equations from here.
You have an extra two equations coming from $f(0) = 2$ and $f(-2) = 0$, meaning that you have four equations in four unknowns, which you can solve accordingly. 
A: The key here is translating these conditions into a system of equations in terms of $a$, $b$, $c$, and $d$. First, you know that $f(0) = 2$ and $f(-2) = 0$, which can be expanded into the equations
\begin{align*}
    d &= 2,\text{ and} \\
   -8a + 4b - 2c + d &= 0
\end{align*}
At the same time, you know that the derivative
$$
f'(x) = 3ax^2 + 2bx + c
$$
of $f(x)$ satisfies $f'(\pm 1) = 0$. Writing this out as a system of equations, you get
\begin{align*}
    3a+2b+c &= 0 \\
    3a-2b+c &= 0.
\end{align*}
Now try to solve for $a,b,c,d$ using all four of these equations. (Really, you should see $d$ immediately and consider three equations instead.)
