Question about a derivative theorem proof

[pc2-brazil]

Good afternoon,

I would like to ask a doubt concerning the proof of a theorem which I found in a Calculus book. The doubt is about a particular detail in the proof.
First, I have to state the definition of relative minimum, which is given at the same source:
"A function f has a relative minimum in c if there is an open interval containing c, where f is defined, such that f(c) ≤ f(x) for all x in this interval."
Now, the theorem and its proof:
If f(x) exists for all values of x in the open interval (a,b) and f has a relative extremum in c, where a < c < b, if f'(c) exists, then f'(c) = 0.
Proof: (For the case in which there is a relative minimum in c.)
If f'(c) exists, then:
$$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$
Since f has a relative minimum in c, by the definition of relative minimum, $$f(c) \leq f(x)$$. So:
$$f(x)-f(c) \geq 0$$
If x approaches c from the right, x - c > 0, therefore:
$$\frac{f(x)-f(c)}{x-c} \geq 0$$ (*)
If this limit exists, then:
$$\lim_{x\to c^+} \frac{f(x)-f(c)}{x-c} \geq 0$$
Analogously, if x approaches c from the left, x - c < 0, therefore:
$$\frac{f(x)-f(c)}{x-c} \leq 0$$ (*)
such that, if this limit exists:
$$\lim_{x\to c^-} \frac{f(x)-f(c)}{x-c} \leq 0$$
Since f'(c) exists, the limits of the above inequalities must equal, and both have to equal f'(c). Then, we have:
$$f'(c) \geq 0$$
and
$$f'(c) \leq 0$$
Since the two inequalities above are true, we conclude that:
$$f'(c) = 0$$
as we wished to demonstrate.
My doubt is this: the definition says that, if there is a relative minimum in c, then f(c) ≤ f(x). I imagine that the "equal" in "less or equal than" part comes from the fact that this definition holds "for all x in this interval", including c. But the text above the parts that I marked with (*) states that "x approaches c from the left" or "x approaches c from the right". Then, since this x is not equal to c, it only approaches c, shouldn't it be possible to write the asterisked parts with "less than" and "greater than", without the "or equal" part? That is:
$$\frac{f(x)-f(c)}{x-c} < 0$$ and
$$\frac{f(x)-f(c)}{x-c} > 0$$
instead of:
$$\frac{f(x)-f(c)}{x-c} \leq 0$$ and
$$\frac{f(x)-f(c)}{x-c} \geq 0$$
It seems to be reasonable, since x approaches c, and c is a relative minimum. But it can't be done, because it would make the proof senseless.
What is wrong with this reasoning?

Thank you in advance.

 

[micromass]

Hi pc2-brazil!

What if f is a constant function, i.e. what if f(x)=3 for all x? Then

$$\frac{f(x)-f(c)}{x-c}=\frac{3-3}{x-c}=0$$

So we need the "or equal" part here.

 

[pc2-brazil]

Hi pc2-brazil!

What if f is a constant function, i.e. what if f(x)=3 for all x? Then

$$\frac{f(x)-f(c)}{x-c}=\frac{3-3}{x-c}=0$$

So we need the "or equal" part here.

OK, thank you, I get it now. I assumed that, for a function to have a relative minimum (c,f(c)) in an interval, it should have only one value c which is smaller than every other value in the interval.