determine wheather each function is even, odd, or neither.?

Question

1. f(x)=x^3-x+1
2. g(x)=9-x^2
3. h(x)=x^3-x/x^4+1

Answer 1

Even means f(x) = f(-x) and odd means f(x) = -f(-x), so try plugging in a positive and negative number (same numeral) and see what you get

2^3 - 2 + 1 = 7
(-2)^3 - (-2) + 1 = -8 + 2 + 1 = -5 so neither

9 - 2^2 = 5
9 - (-2)^2 = 5 so even

Like that

Answer 2

1. Is Odd (it has a odd (3) power)
2. Is even (it has a even (2) power)
3. Is odd (because x^3/x^4 = 1/x and x has odd (1) power)

EDIT:
3) I believe it's neither.

Answer 3

weird and wonderful potential f(-x) = -f(x) Even potential f(x) = f(-x) i) look at f(x), f(-x), and -f(x) and notice which relationships above are fulfilled. f(x) = 6x^5 - 5x -f(x) = -6x^5 f(-x) = 6(-x)^5 - 5(-x) = -6x^5 + 5x f(-x) = -f(x) so it truly is weird and wonderful. f(x) does no longer equivalent f(-x) so it is not even. ii) g(x) = 4x^2 + 2x -g(x) = -4x^2 - 2x g(-x) = 4(-x)^2 + 2(-x) = 4x^2 - 2 g(-x) does no longer equivalent -g(x) so it is not weird and wonderful. g(x) does no longer equivalent g(-x) so it is not even. iii) i will make it easier to try this one your self. word that every time x is raised to easily even powers, the function is even (as a results of fact x^2 = (-x)^2 case in point). whilst x is raised to easily weird and wonderful powers, the function is weird and wonderful (as a results of fact (-x)^3 = -(x^3) case in point). whilst there are the two even and weird and wonderful powers, it truly is neither even nor weird and wonderful.

Answer 4

1)neither
2)even
39odd