Question

Here is a simple one. Most Calculus texts will have a proof.

Suppose \(f'(x) = g'(x)\) for all \(x \in \mathbb{R}\).
Prove that \(f(x) - g(x) = c\) with some constant \(c \in \mathbb{R}\) for all \(x\ in \mathbb{R}\)

Answer 1

\[d/dx ( f- g) = d/dx (c) \]
\[f' - g' = 0 \]
add g' to both sides

\[f' = g' \]

Answer 2

We know the derivatives are the same, you are trying to prove that this implies that the difference between the functions is a constant function, not the other way around.

Answer 3

Hint: MVT

Answer 4

\[f'(x) = \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}, g'(x) = \lim_{h\rightarrow0}\frac{g(x+h)-g(x)}{h}\]
\[f'(x)-g'(x) = 0 \iff\]
\[\lim_{h\rightarrow0}\frac{[f(x+h)-g(x+h)]-[f(x)-g(x)]}{h} = 0\]

would this work at all?

Answer 5

you are working way too hard

Answer 6

i know <.< limited knowledge of this is showing unfortunately =/
anywhos, would it work? lol

Answer 7

\[h(x) = f(x) - g(x)\]\[h'(x) = f'(x) - g'(x) = 0\]
Work from here

Answer 8

You should be using MVT somewhere in the proof.

Answer 9

If h(x) is not a constant function, how would this contradict the fact that h'(x) = 0, using the mean value theorem.

Answer 10

If \(h(x)\) is not constant then there is some \(x, y\) such that \(h(x) \neq h(y)\)

Answer 11

and then...
\[\frac{h(x) - h(y)}{x-y}\]
wouldnt be equal to 0?
im trying to see this lol >.<

Answer 12

Yep and then that contradicts the fact that \(f'(x) = 0\). There would be some \(c\) such that \(f'(c) \neq 0\)

Answer 13

right right. i believe i see it now.

Answer 14

That's all you need to show. Therefore h(x) is constant.

Answer 15

interesting indeed! thank you for putting up with my sluggishness >.<
you too zarkon, thank you.

Answer 16

MVT is a powerful tool

Answer 17

ive taken one Real Analysis class so far, and the professor skipped over it. it wasnt like we did reach that chapter or section. He just skipped it completely.

Answer 18

didnt*

Answer 19

That's really strange

Answer 20

he was a horrible professor.

Answer 21

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e45d1fe0b8b3609c724342a

Answer 22

i hated that class so much. im gonna try and retake it with someone else.

Answer 23

You can probably cover what you've missed in that class with a good text. I recommend Spivak's text on Calculus.

Answer 24

http://www.amazon.ca/Calculus-Michael-Spivak/dp/0914098896

Answer 25

http://www.amazon.com/Calculus-4th-Michael-Spivak/dp/0914098918/ref=sr_1_1?ie=UTF8&qid=1313199043&sr=8-1
?

Answer 26

oh you posted it lol

Answer 27

doesnt sound like a bad idea, i'll buy it once my school money comes in :)