Question

Prove that f : R → R where f(x) = |x| is neither injective nor surjective.

Answer 1

f(x)=|x|

f(-1)=|-1|=1

f(1)=|1|=1

since different elements 1 and -1 have same image 1.So, |x| is not injective function(one-one function).

SECOND METHOD TO CHECK INJECTIVITY:Graphical approach

if any line parallel to x-axis cut the given curve at maximum 1 point,then function is injective.But,here line parallel to x-axis cut the curve at 2 points.hence,|x| is not injective.

METHOD TO CHECK SURJECTIVITY:

f(x)=|x|,f:R->R

we have to determine the range of |x|..

f(-4)=|-4|=4

f(4)=|4|=4

we can also identify from the graph that for every value of x,f(x) always >=0.

Range of f(x)=[0,infinity)

but co-domain given=R, i.e.(-infinite,infinite)

Range of f(x) is not equal to co-domain of f(x).Hence |x| is not surjective function.

Hence,proved!