How do you determine if # h(x)= x^7+x^3+7# is an even or odd function?

Question

Lincoln Anderson · Accepted Answer

It is neither.

The shortcut method is to note that #h(x)# is a polynomial with terms with a mixture of odd degree (#x^7# and #x^3#) and even degree (#7#), so it is neither an odd nor even function. An even function satisfies #f(-x) = f(x)# for all #x# in the domain. An odd function satisfies #f(-x) = -f(x)# for all #x# in the domain. We find: #h(1) = 1+1+7=9# #h(-1) = -1-1+7 = 5# So #h(x)# satisfies neither condition. In general, one effective method to tell whether a function #f(x)# is odd, even or neither is to substitute #-x# for #x# into its definition and see if the result simplifies to #f(x)# or #-f(x)#.

Sadie Ackerman · Answer

undefined To determine whether the function $ h(x) = x^7 + x^3 + 7 $ is even or odd:

1. **Even Function:** If $ h(-x) = h(x) $ for all $ x $ in the domain of $ h(x) $, then $ h(x) $ is even.

2. **Odd Function:** If $ h(-x) = -h(x) $ for all $ x $ in the domain of $ h(x) $, then $ h(x) $ is odd.

For $ h(x) = x^7 + x^3 + 7 $:

- **Even Function Test:** 
  $$ h(-x) = (-x)^7 + (-x)^3 + 7 = x^7 - x^3 + 7 $$
  Since $ h(-x) $ is not equal to $ h(x) $, $ h(x) $ is not even.

- **Odd Function Test:** 
  $$ h(-x) = (-x)^7 + (-x)^3 + 7 = -x^7 - x^3 + 7 $$
  Since $ h(-x) $ is not equal to $ -h(x) $, $ h(x) $ is not odd.

Therefore, $ h(x) = x^7 + x^3 + 7 $ is neither even nor odd.