Let f : R → R be any function. Define g : R → R by g(x) = |f(x)| for x → R. Then, g is

continuous if f is continuous

Let h(x) = |x|. Then, h : R → R is continuous many-one and into function.

We have,

hof(x) = h(f(x)) = |f(x)| = g(x)

Since composition of continuous functions is continuous.

Therefore, g(x) is continuous if f is continuous.

Since, composition of two bijections is a bijection. Here, h(x) is many-one. So, g(x) cannot be one-one even if f is one-one. Also, g(x) cannot be onto even if f is onto.

We observe that f(x) = sin x is everywhere differentiable but |sin x| is not differentiable at x = nπ, n ∈ Z. Therefore, g(x) need not be differentiable even if f is differentiable.