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 !!!
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$\begingroup$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 $f'(c)=k$.
$\endgroup$ 5 $\begingroup$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.
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