Question

Consider the function f(x) = round(x), which rounds the input, x, to the nearest integer. Is this function one-to-one? Explain or justify your answer.

Answer 1

Answer:

No, it's not a one-to-one function

Step-by-step explanation:

Given

$f(x) = round(x)$

Required

Determine if this function is one-to-one

I'll answer this question using the following illustrations;

Assume x = 4.6

$f(x) = round(x)$ would be

$f(4.6) = round(4.6)$

$f(4.6) = 5$

Also assume x = 4.8

$f(x) = round(x)$ would be

$f(4.8) = round(4.8)$

$f(4.8) = 5$

More generally;

$4.5 \leq\ x\leq\5.4$

Would result in:

$f(x) = 5$

Since, there's a always a possibility of f(x) having one value for various values of input x, then we can conclude that the function is not a one-to-one function