It was told to me an exercise:
Represent a continuous and differentiable function whose derivative is annulled in points $A(-1, 4)$ and $B(2, -3)$ and that complies these conditions:
$\displaystyle\lim_{x\to-\infty} f(x) = -\infty\qquad \lim_{x\to+\infty} f(x) = +\infty$
But I found it very easy and I wanted to go further. My question is: is there any to obtain the function $f(x)$?
I've thought that:
To simplify the problem, I considere that $f'(x)$ is only annulled in points $A$ and $B$, so $f'(x)$ is a 2nd grade equation.
Due to that, $f(x)$ has to be a 3rd grade equation, so I say: $f(x)= ax^3+bx^2+cx+d, \qquad a,b,c,d\in \mathbb R$.
Since $f'(x)$ is the derivative of $f(x)$, I've got that $f'(x) =3ax^2+2bx +c$.
Then, I can say the following:
Because $ A,B\in f\longrightarrow f(-1)=4$ and $f(2)=-3$.
And because $A$ and $B$ nullify the derivative of $f(x) \longrightarrow f'(-1)=0$ and $f'(2)=0$.
If I pose the equations of that supositions, I've got that:
$$ \left. \begin{array}{rl}4 &= a(-1)^3 + b(-1)^2 + c(-1) + d\\ -3 &= a · 2^3 + b · 2^2 + c · 2 +d\\ 0&=3a(-1)^2 +2b(-1)+c\\ 0&=3a · 2^2 + 2b · 2 + c \end{array}\right\} = \left. \begin{array}{rl}4 &= -a + b - c + d\\ -3 &= 8a + 4b + 2c +d\\ 0&=3a - 2b +c\\ 0&=12a + 4b + c \end{array}\right\}$$
but I can't find a way to solve this. Is there any way to get that function (or solve the system with another equation(s))? Have I done something wrong?
