- #1

pc2-brazil

- 205

- 3

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:

[tex]f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}[/tex]

Since f has a relative minimum in c, by the

**definition**of relative minimum, [tex]f(c) \leq f(x)[/tex]. So:

[tex]f(x)-f(c) \geq 0[/tex]

If x approaches c from the right, x - c > 0, therefore:

[tex]\frac{f(x)-f(c)}{x-c} \geq 0[/tex]

**(*)**

If this limit exists, then:

[tex]\lim_{x\to c^+} \frac{f(x)-f(c)}{x-c} \geq 0[/tex]

Analogously, if x approaches c from the left, x - c < 0, therefore:

[tex]\frac{f(x)-f(c)}{x-c} \leq 0[/tex]

**(*)**

such that, if this limit exists:

[tex]\lim_{x\to c^-} \frac{f(x)-f(c)}{x-c} \leq 0[/tex]

Since f'(c) exists, the limits of the above inequalities must equal, and both have to equal f'(c). Then, we have:

[tex]f'(c) \geq 0[/tex]

and

[tex]f'(c) \leq 0[/tex]

Since the two inequalities above are true, we conclude that:

[tex]f'(c) = 0[/tex]

**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:

[tex]\frac{f(x)-f(c)}{x-c} < 0[/tex] and

[tex]\frac{f(x)-f(c)}{x-c} > 0[/tex]

instead of:

[tex]\frac{f(x)-f(c)}{x-c} \leq 0[/tex] and

[tex]\frac{f(x)-f(c)}{x-c} \geq 0[/tex]

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.