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?
Best Answer
Hard to know, because of inherent ambiguities in reading a graph. But it is clear that it is intended that there is mirror symmetry across the origin, that is $f(-x)=-f(x)$. So our cubic $P(x)$, if it is a cubic, has equation of the form $P(x)=ax^3+bx$.
The local extrema seem to be at $x=\pm 1$, so $P'(1)=0$. Since $P'(x)=3ax^2+b$, we get $3a+b=0$. (We will get no further information from considering $x=-1$, since the symmetry is built into the shape $ax^3+bx$.)
It also looks as if the value of $P(x)$ at $x=1$ is $-2$. Thus $a+b=-2$.
Solve the system $3a+b=0$, $a+b=-2$. We get $a=1$ and $b=-3$.
So from the information we have used, it is plausible to think that $P(x)=x^3-3x$. Let's make a partial check. We have $x^3-3x=0$ at $x=0$ and $x=\pm\sqrt{3}$. The picture is not inconsistent with these values.