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

[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

 

[fzero]

$$f(b) \ - \ f(a) \ = \ f'(c)(b \ - \ a)$$

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.

 

[sponsoredwalk]

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.