If $a$ and $b$ are consecutive roots of $f$, show that $f'(a)\cdot f'(b)\leq 0.$

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}\ $ be a differentiable and continuous function. If $a$ and $b$ are consecutive roots of $f$, show that $f'(a)\cdot f'(b)\leq 0.$

I think that I have to define an specific function to reach the result, but today I can't think too much :'(

Can someone please explain this?

Answer 1

I think it is easier to work by contraposition.

Assume $f'(a)f'(b)>0$, WLOG we can assume both are positive (else consider $-f$ instead of $f$), by differentiability there exist $f(a+h)>f'(a)h>0$ and $f(b-h)<-f'(b)h<0$ and there exist a $c$ inside $[a+h,b-h]$ such $f(c)=0$ by IVT.

Answer 2

Hint: The word consecutive plays an important role here. Try to sketch such a function having two consecutive roots and notice that the function is decreasing in some neighborhood of the first root and increasing in some neighborhood of another. This means derivatives have different signs. Also, try to figure out when the equality holds.

Answer 3

Suppose $f$ has consecutive zeros at $a<b.$ We then have $f(x)\ne 0$ in $(a,b).$ Because $f$ is continuous, we can apply the IVT to see one of the following must hold: i) $f>0$ on $(a,b);$ ii) $f<0$ on $(a,b).$

WLOG, i) holds. Then for $x\in (a,b),$

$$\tag 1 \frac{f(x)-f(a)}{x-a}>0.$$

The limit of $(1)$ as $x\to 0^+,$ which is $f'(a),$ is therefore $\ge 0.$

The same kind of argument shows $f'(b)\le 0.$

The product of a nonnegative number and a nonpositive number is nonpositive. That shows $f'(a)f'(b)\le 0$ as desired.