# Darboux Theorem

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 !!!

• 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

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 $f'(c)=k$.
• 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
• @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
• $F^{\prime}(a) <0$ 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 3:32