Intermediate value property for derivatives I'm going over a proof of a special case of the Intermediate value theorem for derivatives. 

It says: 


  Consider $a, b\in I$ with $a&lt;b$.
  
  Suppose first that $f'(a)&lt;0&lt;f'(b)$. We will show $\exists x\in(a, b)$
  such that $f'(x)=0$. 
  
  Since $f'$ exists on $[a, b]$, we must have $f$ continuous on $[a, b]$, so the  Extreme Value Theorem implies that $f$ attains its minimum
  at some point $x\in[a, b]$. This minimum point cannot be an endpoint
  of $[a, b]$.


I just don't understand this last sentence. Why can't the minimum point be an endpoint of $[a, b]$?
Since $$f'(a)=\lim\limits_{h\rightarrow0^+} {f(a+h)-f(a)\over h }&lt;0,$$ it follows that  ${f(a+h)-f(a)\over h }&lt;0$ for $h&gt;0$ sufficiently small (with $a+h&lt;b$). From this, it follows that  ${f(a+h)-f(a)  }&lt;0$ for $h&gt;0$ sufficiently small; thus, the minimum can't occur at $x=a$.

A similar argument will show the minimum can't occur at $x=b$.
Since $$f'(a)=\lim\limits_{h\rightarrow0^+} {f(a+h)-f(a)\over h }&lt;0,$$ it follows that  ${f(a+h)-f(a)\over h }&lt;0$ for $h&gt;0$ sufficiently small (with $a+h&lt;b$). From this, it follows that  ${f(a+h)-f(a)  }&lt;0$ for $h&gt;0$ sufficiently small; thus, the minimum can't occur at $x=a$.

A similar argument will show the minimum can't occur at $x=b$.