I'm trying to find a function that is even, end behavior goes towards negative infinity, has maxes only at (-3,5), (0,5), and (3,5), and has minimums only at (-1,-5) and (1,-5).
I tried a system of equations that satisfies f(x)=$ax^8$+$bx^6$+$cx^4$+$dx^2$+5 and f'(x)=$8ax^7$+$6bx^5$+$4c^3$+$2dx$ but I got this, which has the points, but mins instead of maxes at x=±3 and an extra max between ±1 and ±3. I think this is because the degree of the function should be 6, but it seems impossible to find with all the requirements. I was suggested to use the sign of the second derivative, but I don't know how to solve a system of equations with only some of them inequalities, and it would likely make the function more complicated due to the degree being 10 or 12.
All help appreciated. Sorry if the tags are off.
You are correct that the degree should be six and all the terms should have even degree. You are also correct that the constant term should be $5$. The pedestrian approach is to write $f(x)=ax^6+bx^4+cx^2+5$ and write a set of simultaneous equations from the points you know. $$729a+81b+9c+5=5\\a+b+c+5=-5\\6\cdot 3^5a+4\cdot 3^3b+2\cdot 3c=0\\ 6a+4b+2c=0$$
Another approach is to work with $f(x)-5$. You know it has roots at $-3,0,3$ which gives you three factors. Then note that $f(1)-5=-10$ and $f'(1)=0$