# Decide if the functions are odd, even or neither: a) f(x)=x^5+x b) f(x)=1-x^4

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The function is even iff, f(x)=f(-x)

The function is odd iff, f(-x)=-f(x)

Let us start with the first function:

a. f(x)=x^5+x

==> f(-x)= (-x)^5-x= -x^5-x= -(x^5+x)=- f(x) then the function is odd

b. f(x)= 1-x^4

==>f(-x)= 1-(-x)^4= 1-x^4 =f(x) then the function is even

First, let's remind what is an odd function:

If f(x) satisfies f(-x)=-f(x), for each value of x, from the domain of the function, then the function is odd.

Now, let's recall what is an even function:

If f(x) satisfies f(-x)=f(x), for each value of x, from the domain of the function, then the function is even.

Based on these rules, let's analyze each function.

Let's start with the function from a), namely f(x)=x^5+x.

We'll check the rule, substituting x by (-x).

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

f(-x) = (-1)^5*(x)^5 - x

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

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

f(-x) = -f(x)

So, f(x)=x^5+x is an odd function

b) f(-x) = 1-(-x)^4

f(-x) = 1-(-1)^4*x^4

f(-x) = 1-x^4

f(-x) = f(x)

So, f(x) = 1-x^4 is an even function.

To decide the functions which are odd.

a) f(x) = x^5+x

b) f(x) 1-x^4.

Solution:

A function f(x) is said to be even if it it satisfies the criteria,

f(x) = f(-x). Or the function is symmetric about y axis.

Also if f(-x) = -f(x), then the function is said to be odd.

a) f(x) = x^5+x.

Replace x by -x in the function.

f(-x) = (-x)^5 +(-x) = - x^5-x = -(x^5+x) = -f(x). So f(-x) = f(x) implies the function is odd.

b) f(x) 1-x^4. Replacing x by -x , we get:

f(-x) = 1-(-x)^4 = 1+x^4 = f(x). So

f(-x) = f(x). So the function as per the crirteria, is an even function and is symmetric around x.