Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
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

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.