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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
    $f(-1)=-a_3-a_1=2$
    $f(-2)=a_3+a_1=-2$
    $f'(-1)=3a_3-2a_1=0$
    $f'(1)=3a_3+2a_1=0$

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?

  • 2
    You have missing coefficients in the derivative expressions. E.g. derivative of $a_3 x^3$ is $3a_3 x^2$, not $a_3 x^2$.2012-08-24
  • 0
    Thank you! It's like I forgot eveyrything during the holidays.2012-08-24
  • 0
    I added the missing coefficients to the equation, but the derivatives are still false.2012-08-24
  • 0
    You turned $a_1$ into $2a_1$ in the expression for $f'(x)$ going from step 2 to step 3.2012-08-24

2 Answers 2