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KLscilevothma
Jul13-03, 06:13 PM
It isn't a homework problem. [:)]
Let f:[rr]--> [rr] be a function such that f(xy)=f(x)+f(y) for all x,y belong to real numbers
a) Show that for all integers n, f(xn)=nf(x)
b) Suppose the inverse of f exists. Show that for all integers n, [f-1(x)]n=f-1(nx)

I know how to do part (a) but not part (b). Could someone please give me some hints/solution. I can't do problems involving "inverse of a function" unless they are very simple, like to find f-1 of f(x)=(1-x)/(1+x) where x does not equal to 1 or -1

Dx
Jul13-03, 06:33 PM
Originally posted by KL Kam
It isn't a homework problem. [:)]
Let f:[rr]--> [rr] be a function such that f(xy)=f(x)+f(y) for all x,y belong to real numbers
a) Show that for all integers n, f(xn)=nf(x)
b) Suppose the inverse of f exists. Show that for all integers n, [f-1(x)]n=f-1(nx)

I know how to do part (a) but not part (b). Could someone please give me some hints/solution. I can't do problems involving "inverse of a function" unless they are very simple, like to find f-1 of f(x)=(1-x)/(1+x) where x does not equal to 1 or -1

a) thats the power rule formula for solving dx
b) this is antidx. remember its the opposite of dx. if your good with the power rule just do it backwards and subtract.

A good toolkit (http://math.vanderbilt.edu/~pscrooke/toolkit.shtml) for you to use.
dx [;)]

mathman
Jul13-03, 06:43 PM
Let x=f(y), or y=f-1(x)
f(yn)=nf(y)=nx
Take inverse on both sides and get
yn=(f-1(x))n=f-1(nx)