The question is the following
Polynomial function of 4th degree crosses the $x$-axis at $x=1$ and $x=-1$. It also has a minimum at $(0,-1/2)$. Determine the function.
So I know that the function has to look like this: $f(x)=ax^4+bx^3+cx^2+dx+e$. I am given the points $(-1,0)$, $(1,0)$ and I can also use the first derivative and equal it to $0$, since there is a minimum there. I still can't figure out the solution. What am I missing?
From the fact it is 4th degree and has roots $1,-1$ we know that the function can be written as
$f(x)=(x-1)(x+1)(ax^2+bx+c)$
From that it has value of $f(0)=-\frac{1}{2}$ we can learn then that $c$ in the above is equal to $\frac{1}{2}$, so we know it is of the form
$f(x)=(x-1)(x+1)(ax^2+bx+\frac{1}{2}) = (x^2-1)(ax^2+bx+\frac{1}{2})=ax^4+bx^3+\frac{1}{2}x^2-ax^2-bx-\frac{1}{2}$
We then can use that $f(0)$ is a minimum, noting that $f'(0)=0$ to learn that $b=0$ since $f'(x) = 4ax^3+3bx^2+x-2ax-b$ and so $f'(0)=-b=0$
Lastly, since this is a minimum and not a maximum, we can learn something about the range of values that $a$ can take, though we can not pinpoint what exactly $a$ is. For $0$ to have been a minimum of the graph, that implies that $f''(0)$ would need to be positive implying that $1-2a>0$ and so $a<\frac{1}{2}$
Final answer:
$$f(x)=(x-1)(x+1)(ax^2+\frac{1}{2})~~~\text{with }a<\frac{1}{2}$$
or written expanded out as
$$f(x)=ax^4-ax^2+\frac{1}{2}x^2-\frac{1}{2}~~~\text{with }a<\frac{1}{2}$$