Question about a derivative theorem proof

In summary: But since a constant function always has the same value for every x, f(x) would be smaller than any other value in the interval, making the relative minimum meaningless.In summary, the theorem states that if a function has a relative extremum (c,f(c)), and if there is an open interval containing c, then f'(c) = 0. However, if f is a constant function, then f(x) is always the same, and so the relative extremum and the open interval do not have any meaning.
  • #1
pc2-brazil
205
3
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:
[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.
 
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  • #2
Hi pc2-brazil! :smile:

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

[tex]\frac{f(x)-f(c)}{x-c}=\frac{3-3}{x-c}=0[/tex]

So we need the "or equal" part here.
 
  • #3
micromass said:
Hi pc2-brazil! :smile:

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

[tex]\frac{f(x)-f(c)}{x-c}=\frac{3-3}{x-c}=0[/tex]

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.
 

Related to Question about a derivative theorem proof

1. What is a derivative theorem?

A derivative theorem is a mathematical rule or principle that explains the relationship between the derivative of a function and the original function. It helps in finding the derivative of more complex functions by using simpler functions as building blocks.

2. What is the purpose of proving a derivative theorem?

The purpose of proving a derivative theorem is to establish a mathematical truth or rule that can be used to solve problems involving derivatives. It also helps in understanding the underlying principles and reasoning behind the derivative concept.

3. How is a derivative theorem proven?

A derivative theorem is typically proven using mathematical induction, which involves proving that the theorem holds for a base case and then showing that it also holds for the next case. This process is repeated until the theorem is proven for all cases.

4. What are some common derivative theorems?

Some common derivative theorems include the power rule, product rule, quotient rule, chain rule, and the derivative of trigonometric functions. These theorems are essential in finding the derivative of various types of functions.

5. How can I apply derivative theorems in real-world situations?

Derivative theorems can be applied in various real-world situations, such as in physics, engineering, economics, and statistics. They can be used to model and analyze changes in quantities over time, calculate rates of change, and optimize functions.

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