1
$\begingroup$

I have a homework Question to answer which is:

True or False: Between 2 sequential roots of f'(x) there is at most one root for $f(x)$

I think this is true since $f(x)$ would be monotone unless f'(x) is equal to $0$ on an interval - but in that case what are 2 sequential values of the roots?

Can someone help me with this question?

Thanks :)

  • 0
    Not true if it isn't differentiable at all values on the interval, even if continuous. But true if differentiable everywhere in the interval.2012-01-06

4 Answers 4

2

Suppose $f$ has two roots $x=a$ and $x=b$. That is $f(a)=f(b)=0$. Assuming f' exists on $[a,b]$, by the Mean Value Theorem (or Rolle's Theorem if you prefer), there is a $c\in(a,b)$ with f'(c)=0.

This implies that if $c$ and $d$ are sequential roots of f' and if f' exists on $[c,d]$, then $f$ has at most one root between $c$ and $d$.


Added:

I am interpreting "between two sequential roots of $f'\ $" to mean that there is a $c$ and a $d$ with f'(c)=f'(d)=0 and f'(x) is non-zero for all $x$ between $c$ and $d$.

  • 0
    Oh OK, That cleats it up. Thanks :)2012-01-06
2

When f' is not continuous then very strange things may occur. So let's assume that f' is continuous, that f'(a)=f'(b)=0 and that f'(x)\ne0 for $a. Then f'(x)>0 for all $x\in\ ]a,b[\ $ or f'(x)<0 for all $x\in\ ]a,b[\ $. Assuming the former it follows that $f$ is strictly increasing on $[a,b]$, so there is at most one zero of $f$ in this interval.

0

Hint: http://en.wikipedia.org/wiki/Mean_value_theorem

  • 0
    Hmm I was thinking of using this somehow - just have not quite figured it out yet, good to see I'm in the right direction though. Do Feel free to hint some more :P2012-01-06
0

It must be true, because the root of f'(x) means it is maximum or minimum of $f(x)$, so the function in this interval is whether decreasing or rising, therefore there can't be more than one root of $f(x)$

Indeed, it can be a curving point. But even if it is a curving point, there is only 1 root in this interval.

  • 0
    @TonyK :) Thanks, I am translating these terms as I am not learning in English - but thanks for the extra knowledge2012-01-06