How do you determine if #f(x)=x^5-x+5# is an even or odd function?

Question

Elias Ackerman · Accepted Answer

#f(x)# is neither even nor odd.

An even function is one for which #f(-x) = f(x)# for all #x# in the domain. An odd function is one for which #f(-x) = -f(x)# for all #x# in the domain. There is a fast way to determine if a polynomial is odd or even: It is strange if every term has an odd degree. It is even if every term has an equal degree. If not, it is neither even nor strange. In our example, #x^5# and #-x# are of odd degree and the constant #5# is of even (#0#) degree. So #f(x)# is neither odd nor even. In particular, we discover: #f(2) = 32-2+5 = 35# #f(-2) = -32+2+5 = -25# So #f(-2)# is neither #f(2)# nor #-f(2)#.

Elias Ackerman · Answer

undefined To determine if $ f(x) = x^5 - x + 5 $ is an even or odd function, we need to examine two properties:

1. Even functions satisfy the condition $ f(-x) = f(x) $ for all $ x $ in the function's domain.
2. Odd functions satisfy the condition $ f(-x) = -f(x) $ for all $ x $ in the function's domain.

First, let's check if $ f(-x) = f(x) $:

$$ f(-x) = (-x)^5 - (-x) + 5 = -x^5 + x + 5 $$

Now, let's compare $ f(-x) $ with $ f(x) $:

$$ f(x) = x^5 - x + 5 $$

Since $ f(-x) $ is not equal to $ f(x) $, the function is not even.

Next, let's check if $ f(-x) = -f(x) $:

$$ f(-x) = -(-x)^5 - (-x) + 5 = -(-x^5) + x + 5 = x^5 + x + 5 $$

Now, let's compare $ f(-x) $ with $ -f(x) $:

$$ -f(x) = -(x^5 - x + 5) = -x^5 + x - 5 $$

Since $ f(-x) $ is not equal to $ -f(x) $, the function is not odd.

Therefore, $ f(x) = x^5 - x + 5 $ is neither an even nor an odd function.