Whether f is a function from Z to R if a) f(n)=±n. b) f(n)=n2+1. c) f(n)=1n2−4.?

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Whether f is a function from Z to R if a) f(n)=±n. b) f(n)=n2+1. c) f(n)=1n2−4.?

rangiranitlh9 · Accepted Answer

Let&#039;s examine each function to determine if it is a function from Z (integers) to R (real numbers).a) f(n)=±n:This function is not well-defined since it does not specify whether it should output +n or −n for each input n. For example, if we take n=1, it is unclear whether f(1) should be +1 or −1. Therefore, function f is not a function from Z to R.b) f(n)=n2+1:For any integer n, the expression n2+1 is always a non-negative number. The square root of a non-negative real number is a real number. Therefore, the function f(n) is well-defined for all integers n, and it is a function from Z to R.c) f(n)=1n2−4:This function is not well-defined for certain integers. Specifically, when the denominator n2−4 becomes zero, the function is undefined. In this case, when n=2 or n=−2, the denominator becomes zero, resulting in a division by zero error. Therefore, the function f(n) is not a function from Z to R.In summary:a) The function f(n)=±n is not a function from Z to R.b) The function f(n)=n2+1 is a function from Z to R.c) The function f(n)=1n2−4 is not a function from Z to R.

Determine whether f is a function from Z to R if a) f(n) = \pm n. NS

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