Question

Determine whether\({\bf{f}}\)is a function from\({\bf{Z}}\)to\({\bf{R}}\)if

a)\({\bf{f}}\left( {\bf{n}} \right) = \pm {\bf{n}}.\)

b)\({\bf{f}}\left( {\bf{n}} \right) = \sqrt {{{\bf{n}}^{\bf{2}}} + {\bf{1}}} .\)

c)\({\bf{f}}\left( {\bf{n}} \right) = {\bf{1}}/\left( {{{\bf{n}}^{\bf{2}}} - {\bf{4}}} \right).\)

Short answer

(a) \(f\) is not a function as it has more than one number of value for each of the value of \(n\) .

(b)\(f\)is a function.

(c) \(f\) is a not a function as the function is undefined at \(n = \pm 2\).

Step-by-step solution

Significance of the function

The function mainly defines the relationship amongst one of more variable. The functions are mainly used to formulate physical relationships.

Determination of the first function

The given equation is expressed as:

\(f\left( n \right) = \pm n\)

Here,\(f\)is not described as a function as each and every element of\(n \in Z\)has more than one number of value except the point\(n = 0\). Taking an example if\(n = 2\)then\(f\left( n \right) = \pm 2\). For\(n = 0\), the function should consist of one value.

Thus, \(f\) is not a function as it has more than one number of value for each of the value of \(n\) .

Determination of the second function

The given equation is expressed as:

\(f\left( n \right) = \sqrt {{n^2} + 1} \)

The function\(f\)is mainly defined for all the values\(n \in Z\)and also the function maps each and every element of\(Z\)to only one element of\(R\).

Thus, \(f\) is a function.

Determination of the third function

The given equation is expressed as:

\(f\left( n \right) = 1/\left( {{n^2} - 4} \right)\)

Here, for\(n = \pm 2\), the function is undefined as it is impossible to divide the function by\(0\). Hence,\(f\)is not a function from\(Z\)to the point\(R\), as\( \pm 2 \in Z\).

Thus, \(f\) is a not a function as the function is undefined at \(n = \pm 2\).