# Comparing f(b) - f(a) to f(x) - f(a)

• sponsoredwalk
In summary, the conversation is discussing the relationship between two equations that are purported to be the same thing for a specific x. One person is asking for clarification on whether the equations are actually equivalent, while another is trying to use the equations to prove a theorem. The expert summarizer then provides a step-by-step derivation of the equations and concludes that they are not equivalent, and may not be helpful in proving the theorem.
sponsoredwalk
$$f(b) \ - \ f(a) \ = \ f'(c)(b \ - \ a)$$

I'm just wondering can we write

$$f(b) \ - \ f(a) \ = \ \frac{f(x) \ - \ f(a)}{x \ - \ a} (b \ - \ a)$$

for some unknown x &

$$f(x) \ - \ f(a) \ = \ \frac{f(b) \ - \ f(a)}{b \ - \ a} (x \ - \ a)$$

I mean, are these the same thing for a specific x?I need to know this before I write the proof of Cauchy's Mean Value Theorem to check
whether I've done it right, please let me know

sponsoredwalk said:
$$f(b) \ - \ f(a) \ = \ f'(c)(b \ - \ a)$$

I'm just wondering can we write

$$f(b) \ - \ f(a) \ = \ \frac{f(x) \ - \ f(a)}{x \ - \ a} (b \ - \ a)$$

for some unknown x &

$$f(x) \ - \ f(a) \ = \ \frac{f(b) \ - \ f(a)}{b \ - \ a} (x \ - \ a)$$

I mean, are these the same thing for a specific x?

I need to know this before I write the proof of Cauchy's Mean Value Theorem to check
whether I've done it right, please let me know

Since I don't know the content of your proof, I can't comment on that, but assuming the mean value theorem, we know that there is a $$c\in (a,b)$$ such that the first equation is true and there is a $$d\in (a,x)$$ such that

$$f'(d) = \frac{f(x) \ - \ f(a)}{x \ - \ a} .$$

However it seems like we would need to impose further conditions on f in order that $$d\in (a,b)$$, so the second equation that you write does not follow from the mean value theorem. I can't see that these equations would help you prove the mean value theorem.

Well I was just thinking about it, would they represent the same thing.

Here, here & here the proof's all give you a complicated looking formula and tell you that
this will prove everything, but where did it come from? I tried to derive it here so please
let me know if I've assumed anything

$$1) \ F(x) \rightarrow \ f(b) \ - \ f(a) \ = \ \frac{f(x) \ - \ f(a)}{x \ - \ a}(b \ - \ a)$$

$$2) \ G(x) \rightarrow \ g(b) \ - \ g(a) \ = \ \frac{g(x) \ - \ g(a)}{x \ - \ a}(b \ - \ a)$$

$$3) \ \frac{F(x)}{G(x)} \rightarrow \ \frac{f(b) \ - \ f(a)}{g(b) \ - \ g(a)} \ = \ \frac{f(x) \ - \ f(a)}{g(x) \ - \ g(a)}$$

$$4) [ \ f(b) \ - \ f(a) \ ] \ \cdot \ [ \ g(x) \ - \ g(a) \ ] \ ] \ = \ [ \ f(x) \ - \ f(a) \ ] \ \cdot \ [ \ g(b) \ - \ g(a) \ ]$$

$$5) H(x) \ \rightarrow \ [ \ f(b) \ - \ f(a) \ ] \ \cdot \ [ \ g(x) \ - \ g(a) \ ] \ ] \ - \ [ \ f(x) \ - \ f(a) \ ] \ \cdot \ [ \ g(b) \ - \ g(a) \ ] \ = \ C$$

$$6) \ H ' (x) \ \rightarrow \ [ f(b) \ - \ f(a) ] g'(x) \ - \ f'(x)[ \ g(b) \ - \ g(a)] \ = \ 0$$

$$7) \ H ' (c) \ \rightarrow \ [ f(b) \ - \ f(a) ] g'(c) \ - \ f'(c)[ \ g(b) \ - \ g(a)] \ = \ 0$$

$$8) \ [ f(b) \ - \ f(a) ] g'(c) \ = \ f'(c)[ g(b) \ - \ g(a)]$$

$$9) \ \frac{f(b) \ - \ f(a)}{g(b) \ - \ g(a)} \ = \ \frac{f'(c)}{g'(c)}$$

It seems a bit strange to write the first and second equations with x seeing as it
can only be b in order to have equality, unless I'm mistaken.

## 1. What does "f(b) - f(a) to f(x) - f(a)" mean?

The expression "f(b) - f(a) to f(x) - f(a)" is a way of comparing the difference between the outputs of a function at two different inputs (b and a) to the difference between the output at a fixed input (a) and a variable input (x).

## 2. Why is it important to compare these two expressions?

Comparing the two expressions allows us to analyze the behavior of a function as the input value changes. It helps us understand how the output changes in relation to the input and can provide insights into the overall behavior of the function.

## 3. How do you calculate "f(b) - f(a) to f(x) - f(a)"?

To calculate the expression, you first need to find the output of the function at the inputs b and a and subtract them. Then, find the output at the fixed input a and subtract it from the output at the variable input x. The resulting values can then be compared.

## 4. What can we learn from comparing these two expressions?

Comparing the expressions can tell us about the rate of change of the function. If the two expressions are equal, it means that the function has a constant rate of change. If the two expressions are not equal, it means that the rate of change is not constant, and the function may have a more complex behavior.

## 5. Can this expression be used for any type of function?

Yes, this expression can be used for any type of function, as long as the function is defined at the given inputs. It is a general way of comparing the behavior of a function and can provide useful insights for various types of functions.

### Similar threads

• Calculus and Beyond Homework Help
Replies
2
Views
498
• Calculus and Beyond Homework Help
Replies
1
Views
552
• Calculus and Beyond Homework Help
Replies
1
Views
369
• Calculus and Beyond Homework Help
Replies
4
Views
880
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
927
• Calculus and Beyond Homework Help
Replies
8
Views
538
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
888