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Darboux's Theorem states that" If $f$ is differentiable on $[a,b]$ and if $k$ is a number between $f^{\prime}(a)$ and $f^{\prime}(b)$, then there is at least one point $c\in (a,b)$ such that $f^{\prime}(c) = k$."

Most commonly found proof goes as follows: Suppose that $f^{\prime}(a) < k < f^{\prime}(b)$. Let $F:[a,b]\rightarrow \mathbb{R}$ be defined by $F(x) = f(x) -kx$ so that $F^{\prime}(x) = f^{\prime}(x) -k$. Then $F$ is differentiable on $[a,b]$ because so is the function $f$. We find $F^{\prime}(a) = f^{\prime}(a) -k <0$ and $F^{\prime}(b) = f^{\prime}(b) -k >0$. Note that $F^{\prime}(a) <0$ means $F(t_1)< F(a)$ for some $t_1\in (a,b)$. So, $F$ can not attain its minimum at $x=a$. Also, for $F^{\prime}(b)>0$, we can find $t_2\in (a,b)$ such that $F(b) < F(t_2)$. Thus neither $a$ nor $b$ can be a point where $F$ attains a local maximum or a local minimum. Since $F$ is continuous on $[a,b]$, it must attain its maximum at some point $c\in (a,b)$. This means that $F^{\prime}(c) = 0$ and hence $f^{\prime}(c) = k$ as desired.

My question is how can one conclude that $F^{\prime}(b) > 0$ means $F$ can not attain a maximum at $x=b$, the end point of the interval. Look at the example $f(x) =x^2$ on $[0,1]$, has absolute maximum at $x =1$ though $f^{\prime}(1) =2 >0$. Have I misunderstood any logic here? Please somebody explain this. Thanks !!!

Martin Sleziak
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Mr. MBB
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  • Does the Extreme value property for a continuous function on a closed interval $[a,b]$ ensures of the existence of local extremum ? – Mr. MBB Dec 24 '15 at 21:42

2 Answers2

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You’ve slightly misstated the argument, and the misstatement is the source of your difficulty. We have $F'(a)=f'(a)-k<0$, so $F$ cannot have its absolute minimum on $[a,b]$ at $a$: there is a $t_1\in(a,b)$ such that $F(t_1)<F(a)$. We also have $F'(b)=f'(b)-k>0$, so $F$ cannot have its absolute minimum on $[a,b]$ at $b$, either: there is a $t_2\in(a,b)$ such that $F(t_2)<F(b)$. Thus, it must have its absolute minimum on $[a,b]$ at some $c\in(a,b)$. By Fermat’s theorem $F'(c)=0$, and hence $f'(c)=k$.

Brian M. Scott
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  • M Scott. As far as I know that $F^{\prime} (c) = 0$ is not required for absolute maximum or minimum. I gave an example :$f(x) = x^2, x\in [0,1]$ has absolute maximum at $x =1$ without being $F^{\prime}(1) =0$. – Mr. MBB Dec 24 '15 at 22:07
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    @Mr.MBB: It is required when the absolute max. or min. is in the interior of the interval and the function is differentiable. That’s why it’s important that neither $a$ nor $b$ is the min. – Brian M. Scott Dec 24 '15 at 22:09
  • Thanks professor Scott. I really appreciate your help. – Mr. MBB Dec 24 '15 at 23:45
  • @Mr.MBB: You’re welcome! – Brian M. Scott Dec 24 '15 at 23:46
  • $F^{\prime}(a) <0$ and $F^{\prime}(b)>0$ means $F$ is not monotonic on $[a,b]$. This means that there exists $x<y<z$ all on $[a,b]$ such that(a) $F(x) <F(y)$ and $F(y)>F(z) – Mr. MBB Dec 27 '15 at 03:32
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Here is my favourite proof of Darboux's theorem :

Let $g : x \mapsto f(x)-kx$. Then because $k$ lies between $f'(a)$ and $f'(b)$, one has $$g'(a)g'(b) = (f'(a)-k)(f'(b)-k) < 0$$

Hence, $g$ is not monotonic, and because $g$ is continuous, it is not injective. So by Rolle's theorem, $g'$ vanishes, i.e. there exists $c$ such that $f'(c)=k$, and you are done.

TheSilverDoe
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