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:

  1. Degrees of the function
    • Graph has 3 zero-crossings -> minimum degrees of 3
  2. General equation and derivatives
    • $f(x) = a_3 x^3+a_2 x^2+a_1 x+a_0$
    • $f'(x)=3a_3 x^2+2a_2 x+a_1$
    • $f''(x)=6a_3 x+2a_2$
  3. Reduce based on the symmetry
    • $f(x) = a_3 x^3+a_1 x$
    • $f'(x)=3a_3 x^2+2a_1$
    • $f''(x)=6a_3 x$
  4. Build the linear system based on the zero-crossing and extremes for the Gaussian elimination
    $f(-1)=a_3 (-1)^3+a_1 (-1)=2$
    $f(-2)=a_3 1^3+a_1 1=-2$
    $f'(-1)=3a_3(-1)^2+2a_1 (-1)=0$
    $f'(1)=3a_3 1^2+2a_1 1=0$
  5. Simplifying

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.



What am I missing or doing wrong?

  • 2
    $\begingroup$ You have missing coefficients in the derivative expressions. E.g. derivative of $a_3 x^3$ is $3a_3 x^2$, not $a_3 x^2$. $\endgroup$ – Kartik Audhkhasi Aug 24 '12 at 3:23
  • $\begingroup$ Thank you! It's like I forgot eveyrything during the holidays. $\endgroup$ – buschtoens Aug 24 '12 at 3:41
  • $\begingroup$ I added the missing coefficients to the equation, but the derivatives are still false. $\endgroup$ – buschtoens Aug 24 '12 at 3:53
  • $\begingroup$ You turned $a_1$ into $2a_1$ in the expression for $f'(x)$ going from step 2 to step 3. $\endgroup$ – Rahul Aug 24 '12 at 4:20

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.


The fact that it goes through $(0,0)$ tells you one factor is $x$ (assuming you think it is there, not $(0.0001,0)$. Then it depends on what you want/can read off the graph. If you think the other roots are $\pm \frac 74$ you have an equation: $y=x(x^2-\frac 74)^2$ But maybe the other roots are different: what you linked to has them at $\sqrt 3$, which isn't too different from $\frac 74$. The real problem is that you can't read a graph to perfect accuracy, so there are many choice that are acceptable.


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