- #1
Andrax
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so before i start this is extracted from spivak's calculus proof
now here is the problem the teacher who taught us limits and dervatives for the first time told us not to use f(x)-f(a) instead of f(a+h)-f(a)
so spivak wanted us to prove that if f a one-one function and is continious on an interval and differentiable on f[itex]^{-1}[/itex](b) then f'(b) exists and (f^-1)'(b)=[itex]\frac{1}{f'(f^-1(b))}[/itex](f^-1 = inverse of f )
i did the opposite of what spivak did , spivak started from the definition of f'(b) while i started wtih the definition of f[itex]^{-1}[/itex](b), so is my proof right?
let f(a) = b then f^-1(b)=a so
f'(f[itex]^{-1}[/itex](b) = f'(a) = [itex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a} [/itex]=[itex]\lim_{x \to f^-1(b)}\frac{f(x)-b}{x-f^-1(b)} [/itex]
now when x approches a=f^-1(b) in the domain of f , f(x)approaches f(a)=b in the domain on f-1 and we can set f(x)=x later on
so =[itex]\lim_{f(x) \to b}\frac{f(x)-b}{x-f^-1(b)} [/itex]
replacing f(x) by the new x in the domain of f^-1 and b by f^-1(b) and x by f^-1(x)= a(I hope I'm right in this phase, please correct me if I'm wrong)
=[itex]\lim_{x \to b}\frac{x-b}{f^-1(x)-f^-1(b)} [/itex] = [itex]\frac{1}{(f^-1)'(b)}[/itex]
i've always been confused when changing the x to f(x) : do we also change the function or it remains intact?Limx->af(x) <=>Limf(x)->f(a) [f(f(x))) or f(x)? ] if i made a fault it would be this, thanks for reading.
now here is the problem the teacher who taught us limits and dervatives for the first time told us not to use f(x)-f(a) instead of f(a+h)-f(a)
so spivak wanted us to prove that if f a one-one function and is continious on an interval and differentiable on f[itex]^{-1}[/itex](b) then f'(b) exists and (f^-1)'(b)=[itex]\frac{1}{f'(f^-1(b))}[/itex](f^-1 = inverse of f )
i did the opposite of what spivak did , spivak started from the definition of f'(b) while i started wtih the definition of f[itex]^{-1}[/itex](b), so is my proof right?
let f(a) = b then f^-1(b)=a so
f'(f[itex]^{-1}[/itex](b) = f'(a) = [itex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a} [/itex]=[itex]\lim_{x \to f^-1(b)}\frac{f(x)-b}{x-f^-1(b)} [/itex]
now when x approches a=f^-1(b) in the domain of f , f(x)approaches f(a)=b in the domain on f-1 and we can set f(x)=x later on
so =[itex]\lim_{f(x) \to b}\frac{f(x)-b}{x-f^-1(b)} [/itex]
replacing f(x) by the new x in the domain of f^-1 and b by f^-1(b) and x by f^-1(x)= a(I hope I'm right in this phase, please correct me if I'm wrong)
=[itex]\lim_{x \to b}\frac{x-b}{f^-1(x)-f^-1(b)} [/itex] = [itex]\frac{1}{(f^-1)'(b)}[/itex]
i've always been confused when changing the x to f(x) : do we also change the function or it remains intact?Limx->af(x) <=>Limf(x)->f(a) [f(f(x))) or f(x)? ] if i made a fault it would be this, thanks for reading.