Let f be differentiable on an interval A If fx notequalto 0

Let f be differentiable on an interval A. If f\'(x) notequalto 0 on A, show that f is one-to-one on A. Provide an example to show that the converse statement need not be true.

Solution

Proof by contradiction\'

Let, f not be one to one

SO there are some x,y in A ,x not equal to y so that

f(x)=f(y)

By Mean Value Theorem there is some c in (x,y) , assuming x<y

so that: f\'(c)=f(x)-f(y)=0

But, f\'(x) is non zero on A
So a contradiction

HEnce, f is one to one in A

Converse is

f is one to one then f\'(x) is non zero

This need not be true because injectivity property does not need differentiability ie the function need not be differentiable so the derivative does not exist for all x in A

eg. A=[0,2]

f(x)=x, x=0 to 1

f(x)=2x-1, x=1 to 2

THis is not one to one but not differentiable at x=1