Question

Let a function f : (0, [infinity]) → (0, [infinity]) be defined by f(x) = |1 - 1/x|. Then f is:
(A) not injective but it is surjective (B) injective only (C) neither injective nor surjective (D) both injective as well as surjective

Answer 1

Given:

f : (0, [infinity]) → (0, [infinity]) be defined by f(x) = |1 - 1/x|.

To Find:

Type of function f .

Solution:

Given f(x) = |1 -1/x|  

• From the modulus function its clear that f(x) values are positive.

• Domain of the function is ( 0 , ∞ )  

We can try graphing the function in few steps.

1. Draw graph 1/x
2. Reflect it with respect to x axis .
3. Now shift it up my 1 unit.
4. Now Reflect the part of the graph under x axis to above it with respect to the x axis itself.

Looking at the final figure,

• If we draw a line parallel to x axis at lower values of y, we can see that it cuts the graph at 2 points.
• This means two x values give same y value.
• This means the function is many to one and not one to one.
• Also values of y ranges from 0 to ∞.

Hence the graph is not injective - one to one .

But since its Range coincides with its Co Domain , it is surjective.

Hence the function f is not injective but surjective.