How do you know if #y= x+x^2# is an even or odd function?

1 Answer
Dec 13, 2015

The function is neither even nor odd.

Explanation:

To test if the fuction is an even function, you would replace #x# with #-x#, and see if the resulting equation is the same as the original equation.

First, write out the original equation as a function by replacing #y# with #f(x)#:

#f(x) = x + x^2#

Second, replace #x# with #-x#:

#f(-x) = -x + (-x)^2#

Third, simplify the equation:

#f(-x) = -x + x^2#

Finally, compare it to the original equation:

#f(x) = x + x^2 \ != \ f(-x) = -x + x^2#

Since #f(x) != f(-x)#, the function is not even.

To test if the fuction is an odd function, you would test to see if #f(-x) = -f(x)#. If the two equations are the same, then the function is odd.

First, write out the original equation as a function by replacing #y# with #f(x)#:

#f(x) = x + x^2#

Second, multiply both sides of the equation by #-1#:

#-f(x) = -1(x + x^2)#

Third, simplify the equation:

#-f(x) = -x - x^2#

Finally, compare it to the equation from our even function test:

#f(-x) = -x + x^2 \ != \ -f(x) = -x - x^2#

Since #f(-x) != -f(x)#, the function is not odd.

If the function is not odd or even, then it is neither.