Question

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

Answer 1

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.