Question

Given that `f(x)= x^3-2x^2-3x +4` solve for the even and odd parts of `f(x)`?

Answer 1

`f(x) = f_e(x) + f_o(x)` where:

`{ (f_e(x) = -2x^2+4 " is an even function"), (f_o(x) = x^3-3x " is an odd function") :}`

Explanation:

Given any function `f(x)` that has a domain symmetrical about `0`, we can decompose it into the sum of an even function and an odd function.

Putting:

`{ (f_e(x) = 1/2(f(x)+f(-x))), (f_o(x) = 1/2(f(x)-f(-x))) :}`

we find:

`f_e(-x) = 1/2(f(-x)+f(x)) = 1/2(f(x)+f(-x)) = f_e(x)`

`f_o(-x) = 1/2(f(-x)-f(x)) = -1/2(f(x)-f(-x)) = -f_o(x)`

`f_e(x) + f_o(x) = 1/2(f(x)+color(red)(cancel(color(black)(f(-x))))) + 1/2(f(x)-color(red)(cancel(color(black)(f(-x))))) = f(x)`

If the original `f(x)` is a polynomial, then we find that `f_e(x)` is the sum of the terms of `f(x)` of even degree and `f_o(x)` is the sum of the terms of `f(x)` of odd degree.

Note that any constant term is of `0` and therefore even degree.

So in our example, given:

`f(x) = x^3-2x^2-3x+4`

we find:

`{ (f_e(x) = -2x^2+4), (f_o(x) = x^3-3x) :}`

Here are `f_e(x)` and `f_o(x)` plotted together:

graph{(y-(x^3-3x))(y-(-2x^2+4)) = 0 [-5.21, 5.2, -5.2, 5.2]}