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I am asked to find fast the number of possible inflection points of: $$y=(x-1)(x-2)^2(x-3)^4(x-4)^3$$

I know if the degree of any polynomial is even, its plot starts from the 2th quadrant to 1st quadrant of $\mathbb R^2$. This was what I could do fast. Any Ideas? Thanks.

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The term $(x-3)^4$ does not produce an inflection point at $3$, so there are no more than $6$. I would guess $6$. But it is certainly not a fully verified guess. – André Nicolas Sep 7 '12 at 16:35
@AndréNicolas: Yes. x=3 is just a root for y. 6? – Babak S. Sep 7 '12 at 16:45
If you think about the product rule, $x=3$ is a double root of the second derivative. But there is no inflection oint at $x=3$. (Think $x^4$, it does not change concavity at $0$.) So two "possible" inflection points are missing. – André Nicolas Sep 7 '12 at 16:51
@AndréNicolas: Can we say for these kind of functions, there is exactly one extreme between two consecutive roots? Thanks. – Babak S. Sep 7 '12 at 17:02
There is at least one, but there could be more than one. – André Nicolas Sep 7 '12 at 17:04
up vote 9 down vote accepted

Just imagine what the graph looks like. It starts above the $x$-axis, crosses below at $x=1$, is tangent to the $x$-axis at $x=2$ and $x=3$, and then crosses above at $x=4$, with an inflection point at $(4,0)$. Thinking about the shape, I count:

  • One inflection point between $x=1$ and $x=2$,

  • Two inflection points between $x=2$ and $x=3$,

  • Two inflection points between $x=3$ and $x=4$, and

  • One inflection point at $x=4$.

Thus there are six inflection points. This makes sense -- the second derivative should have eight zeroes, but two of them are at $x=3$, leaving six inflection points.

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Oh yes. I see it now. Thanks Jim Thanks @André. – Babak S. Sep 7 '12 at 16:51

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