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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 $cin (a,b)$ 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 $t_1in (a,b)$. So, $F$ can not attain its minimum at $x=a$. Also, for $F^{prime}(b)>0$, we can find $t_2in (a,b)$ 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 $cin (a,b)$. 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 !!!

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 $t_1in(a,b)$ 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 $t_2in(a,b)$ such that $F(t_2)<F(b)$. Thus, it must have its absolute minimum on $[a,b]$ at some $cin(a,b)$. By Fermat’s theorem $F'(c)=0$, and hence $f'(c)=k$.