Help me understand proof.

  • Thread starter Miike012
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  • #1

Homework Statement

y' = f'(x)
then y = f(x)

Let g(x) be any other integral of y' = f'(x)
That is, g'(x) = f'(x).

Now show that g(x) can differ from f(x) by at most a constant.

Then let w' = f'(x) - g'(x) = 0
Then w as a function of x must be w = constant
Hence g(x) = f(x) + constant.

Are they saying that because
w' = f'(x) - g'(x) = 0
w' = 0
Then because the slope of some curve is always zero, therefore it would be a line parrallel to the x-axis... that makes it a constant?

My real concern is.... how did they assume the g'(x) equals f'(x) ? Because if g(x) =/= f(x) then the difference of w' would be greater than a constant.....???

Answers and Replies

  • #2
I am not 100% sure of this but this my 2 cents.
consider this. f(x)=x+a (a being a real number) then f'(x)=1 now it is given that f'(x)=g'(x) the g'(x)=1 (or any other derivative of any given function. now
seeing that[tex]\int[/tex] ] f'(x) dx =f(x)+c the same applied for g'(x) so there is always just a single constant difference i.e

f(x)=x+1[tex]\Rightarrow[/tex] f'(x)=x=g'(x) [tex]\leftharpoondown[/tex] g(x)=x+4 or what ever.
  • #3
y' = f'(x) and y = f(x)
They state: let g(x) be any other integral of y' = f'(x)
This means g'(x) = y' = f'(x)
so g(x) = int(y')dx
g(x) = int(f'(x))dx
g(x) = f(x) + constant

To clarify your proof:

assign a new function w(x) such that it's derivative w'(x) = f'(x) - g'(x)
but f'(x) = g'(x) [stated above]
so w'(x) = 0
Now since w(x) is generally a function of x, and it's derivative w'(x) = 0, in this case this must mean w(x) = constant
Hence (integrating both sides and gathering up all constants)
constant = f(x) - g(x)
or f(x) = g(x) + constant

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