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

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

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

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.