How do you determine if #f(x)=1/x^2# is an even or odd function?

Question

Elias Ackerman · Accepted Answer

By substituting (-1x) for all #x#'s in the function, then evaluate the function.

To determine if a function is even, odd or neither, plug in (-1#x#) in place of all #x#'s in the function, then determine if any of the signs in the function have changed from positive to negative, negative to positive, or neither. Even function: NO sign changes Odd function: ALL signs change Neither: some signs change, some signs do not change. Example: #f(x)=1/x^2# 1. Replace 'all' #x# 's with (-1x): #f(-1x)=1/(-1x)^2# 2. Check to see if any signs have changed in the function. Since #(-1x)^2=x^2#, the sign of the function has not changed. So this function is an EVEN function.

Quick Trick for advanced students: Even functions: Every #x#-term will have an even exponent, and a constant term may or may not exist. (#f(x)=x^4-x^2+2# is an even function. Odd functions: Every #x#-term will have an odd exponent and a constant term will not exist. #f(x)=x^5+x^3+x# is an odd function. Neither: A combination of both even and odd exponents will exist or at least one odd exponent will exist as well as a a constant term. #f(x)=x^5-x^4# and #f(x)=x^5-x^3-2# are neither even nor odd.

Elias Ackerman · Answer

undefined To determine if $ f(x) = \frac{1}{x^2} $ is an even or odd function, we can use the following criteria:

1. Even function: If $ f(x) = f(-x) $ for all $ x $ in the domain of $ f(x) $, then $ f(x) $ is an even function.

2. Odd function: If $ f(x) = -f(-x) $ for all $ x $ in the domain of $ f(x) $, then $ f(x) $ is an odd function.

For $ f(x) = \frac{1}{x^2} $, we have:

$ f(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = f(x) $

Since $ f(-x) = f(x) $ for all $ x $ in the domain of $ f(x) $, $ f(x) $ is an even function.