Let f : R → R be a function. Define g:R → by g(x) = |f(x)| for all x. Then g is
g(x)=|f(x)|≥0.So g(x) cannot be onto. if f(x) one-one and f(x1)=−f(x2) then
So, 'f(x) is one-one' does not ensure that g(x) is one-one.
If f(x) is continuous for x∈R, (x) is also continuous for x∈R.
This is obvious from the following graphical consideration.
So g is continuous if f is continuous.