Name:
Anonymous
2009-01-16 18:01
It has many uses, but if the derivative of a function is greater than 0 for all x from (-inf,inf) does that mean the function itself is one to one?
Name:
Earth
2009-01-16 19:47
Yes, I'd think the best way to prove it would be to use the mean value theorem to prove the contra positive.
Take our f s.t f'(x) > 0 for all x. If there exists x and y s.t f(x) = f(y) wlog x < y
Then the mean value theorem states that there exists a z in (x,y) s.t f'(z) = [f(x) - f(y)]/x-y = 0 contradiction.
similarly you can show the function must be monotonic increasing if f' > 0, obviously both these only work with a strict inequality.
Name:
Anonymous
2009-01-16 22:38
>>1
of course it does. also works if dy/dx is always < 0