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Let $f: [a,b] \longrightarrow \mathbb{R}( a<b)$, $f$ is continuous and differentiable.

We assume that $f$ and $f'$ are increasing and $f(a)<0, 0<f(b)$.

Show that $f$ has a unique zero which we denote $\lambda$ and prove that $f'(\lambda)>0$

I have used IVT theorem to show that $\lambda$ exists.

*For the uniqueness if $f'$ is continuous, I can use $$f(\mu)-f(\lambda)=\int_\lambda^\mu f'(x) \, \mathrm dx$$

Except that f is only differentiable..

Thank you in advance for your help

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    $\begingroup$ Try to prove $f,f'$ increasing and $f(a) < f(b)$ $\Rightarrow f$ is strictly increasing. Then you're done. $\endgroup$
    – AlexR
    Jan 30 '14 at 17:17
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Assume $f(\lambda)=0$ (you have already show the existence of at least one such $\lambda$ with the IVT). Then by the MVT, $$f(\lambda)-f(a)=(\lambda-a)f'(\xi)$$ for some $\xi$ with $a<\xi<\lambda$. As $f(\lambda)-f(a) = -f(a) >0$ and $\lambda-a\ge 0$ we conclude (because $f'$ is increasing) that $$ f'(\lambda)\ge f'(\xi)>0.$$ Assume $\lambda<\mu$ with $f(\lambda)=f(\mu)=0$. Them the MVT leads (again using that $f'$ is increasing) to the contradiction $$ f(\mu)-f(\lambda)=(\lambda-\mu)f'(\xi)\ge (\lambda-\mu)f'(\lambda)>0.$$

Remark: We did not use that $f$ is increasing.

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    $\begingroup$ Wouldn't it be $f(\lambda)-f(a) \gt 0$ since $f(a) \lt 0$ and $f(\lambda)=0$? $\endgroup$
    – philmcole
    Jan 22 '18 at 15:51
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You don't need the fact that $f$ is increasing, only $f'$.

First, as you said, use IVT for existence.

Second, consider your integral $f(a)-f(0)$ where $f(a)=0$ and conclude that $f'(a)$ must be positive. (Edit: as per your comment, if you don't like integral, use MVT to show the same fact.)

Finally, given this, conclude that for $b>a$: $f(b)=f(b)-f(a)$ must be positive as well.

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  • $\begingroup$ The only one you have in your own solution. :) $\endgroup$
    – Vadim
    Jan 30 '14 at 17:32
  • $\begingroup$ I have added a comment to show alternative way using MVT instead of the integral, if that is something you can use (and I believe it is, because you used IVT). $\endgroup$
    – Vadim
    Jan 30 '14 at 17:34
  • $\begingroup$ See my edit above. $\endgroup$
    – Vadim
    Jan 30 '14 at 17:35
  • $\begingroup$ Because this is how $a$ was chosen. It is any point where $f(a)=0$. Take any such point. Show using integral or MVT (together with the fact that $f'$ is increasing) that $f'(a)>0$. Then use this and again integral or MVT to show that for any $b>a$: $f(b)-f(a)=f(b)-0=f(b)>0$. $\endgroup$
    – Vadim
    Jan 30 '14 at 17:40
  • $\begingroup$ Once you show that, it immediately implies that there could not be two such points. $\endgroup$
    – Vadim
    Jan 30 '14 at 17:41
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A different approach without need of integrals.

Since $f$ is increasing, $f'(x)\ge0$. Since $f(a)<0$ and $f(b)>0$, there is a $t\in(a,b)$ such that $f(t)=0$. Let $\lambda=\inf_{t\in(a,b)}f(t)=0$. By continuity, $f(\lambda)=0$ , and $a<\lambda<b$. If $f'(\lambda)=0$, then $f'(x)=0$ for $x\in(a,\lambda)$. This implies that $f$ is constant in$(a,\lambda)$, a contradiction since $f(a)<0$. Thus $f'(\lambda)>0$. This implies that $f(x)>0$ if $\lambda<x<b$ and $f(x)<0$ if $a<x<\lambda$.

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