Consider this graph of a polynomial function. The correct equation is: $f(x) = x^3 - 3x$
Strictly speaking I don't know the equation and have to find it. So far I've done this:
- Degrees of the function
- Graph has 3 zero-crossings -> minimum degrees of 3
- General equation and derivatives
- $f(x) = a_3 x^3+a_2 x^2+a_1 x+a_0$
- $f'(x)=3a_3 x^2+2a_2 x+a_1$
- $f''(x)=6a_3 x+2a_2$
- Reduce based on the symmetry
- $f(x) = a_3 x^3+a_1 x$
- $f'(x)=3a_3 x^2+2a_1$
- $f''(x)=6a_3 x$
- Build the linear system based on the zero-crossing and extremes for the Gaussian elimination
$f(-1)=a_3 (-1)^3+a_1 (-1)=2$
$f(-2)=a_3 1^3+a_1 1=-2$
$f'(-1)=3a_3(-1)^2+2a_1 (-1)=0$
$f'(1)=3a_3 1^2+2a_1 1=0$ - Simplifying
$f(-1)=-a_3-a_1=2$
$f(-2)=a_3+a_1=-2$
$f'(-1)=3a_3-2a_1=0$
$f'(1)=3a_3+2a_1=0$
I don't know how to continue since $a_1$ and $a_3$ have to be $0$ to match both last equations, which they aren't. If I implement the correct $a_1$ and $a_3$ I get false statements.
$f'(-1)=3×1-2×(-3)=3+6=9≠0$
$f'(1)=3×1+2×(-3)=3-6=-3≠0$
What am I missing or doing wrong?