I want to construct a polynomial of degree at least 4 with local maximums at $(-2,1)$ and $(3,4)$ and local minimum at $(1,-2)$. It's easy to draw to have some idea of how $f$ is.
I've tried to solve a linear system with $f(x)=ax^4+bx^3+cx^2+dx+e$
$f(3)=4, f(1)=-2, f(-2)=1$
But I don't know what to do with $f'(x)$ because there are two variables left to solve the system but three local extremes' conditions to fix: $f'(-2)=0, f'(1)=0, f'(3)=0$.
You're right to be concerned. If you have six conditions and five unknowns, you would have to relax one condition and pray that the missing condition is met "accidentally".
One way around it is that your polynomial has to be at least degree four, so there's nothing stopping you from finding a fifth degree polynomial that meets all of those conditions. You have to be sure to double-check your solution to make sure that the turning points are all maxima or minima as you had hoped, though, because sometimes the gods of polynomial interpolation can be fickle.
Let $$ f'(x)=(ax+b)(x+2)(x-3)(x-1)$$
Upon integration you find a polynomial for $f(x)$ with parameters $a,b,c$ which could be found by the given information about $f(x)$ such as $f(-2)=1$ and so forth.
Hint
This makes it way so harder. As the extrema happen at $\{-2,1,3\}$, then these must stand as the roots of the 1st order differentiation i.e.$$f'(x)=a(x+2)(x-1)(x-3)g(x)$$ where $g(x)$ is another polynomial that can be taken equal to $1$ in its simplest case. For deciding whether the extrema are minimum or maximum, you can use the 2nd order differentiation test and to obtain $f(x)$ integrate $f'(x)$.
P.S.
If the problem had no solution with $g(x)=1$, try a more general form of $g(x)$.
Let $f(x):=ax^4+bx^3+cx^2+dx+e$, so $f'(x)=4ax^3+3bx^2+2cx+d$ and $f''(x)=12ax^2+6bx+2c$
First off, the derivative of $f$ should vanish for $x\in\{-2,1,3\}$, so: $$\begin{cases}-32 a + 12 b - 4 c + d=0\\ 4 a + 3 b + 2 c + d=0\\ 108 a + 27 b + 6 c + d=0\end{cases}$$ The second derivatives should be negative for local maxima and positive for local minima: $$\begin{cases}24 a - 6 b + c<0\\ 54 a + 9 b + c<0\\6 a + 3 b + c>0\end{cases}$$ Now, the additional conditions give: $$\begin{cases}81 a + 27 b + 9 c + 3 d + e = 4\\a + b + c + d + e + 2 = 0\\16 a - 8 b + 4 c - 2 d + e =1 \end{cases}$$
Wolfram Alpha says that no solutions exist. I checked by hand, imposing only 5 out of 6 conditions each time, and if my calculations didn't go extremely wrong, there should indeed be no such polynomials.
$$f(x)=\sum_{i=0}^na_ix^i$$
$$\begin{cases} f(-2)=1\\ f'(-2)=0\\ f''(2)\le0\\ f(3)=4\\ f'(3)=0\\ f''(3)\le0\\ f(1)=-2\\ f'(1)=0\\ f''(1)\ge0\\ n\ge 4 \end{cases}$$
Given there is no upper limit on the degree of the polynomial, there are an infinite number of solutions. There may not be one for $n=4$.
