even and odd functions help thanks its greatly appreciated?

Question

Determine if either the function is even odd or neither than describe its symmetry

f(x) = x^6-2x^2+3

h(x) = x^3-5

g(x) = x^3-5x

Answer 1

f(-x) = (-x)^6 - 2(-x)^2 + 3
= x^6 - 2x^2 + 3
= f(x).
Therefore f is even, and y = f(x) is symmetrical about the y axis.

h(-x) = (-x)^3 - 5
= -x^3 - 5
That is neither h(x) nor -h(x).
Therefore h is neither even nor odd. It is symmetrical about the point (0, -5).

g(-x) = (-x)^3 - 5(-x)
= -x^3 + 5x
= -g(x).
Therefore g is odd, and y = g(x), is symmetrical about the origin.

Answer 2

If it's even, f(-x)= f(x), in which case:
f(-x) = (-x)^6-2(-x)^2+3 = x^6-2x^2+3 = f(x) = even function

h(-x) = (-x)^3-5 = -(x^3)-5 =/= h(x), therefore it's not even

g(-x) = (-x)^3-5(-x) = -(x^3)+5x =/= g(x), therefore it's not even


For a function to be odd, f(-x) = -f(x)
-h(x) = -(x^3-5) = -(x^3)+5
When solving for evens, we found that h(-x)=-(x^3)-5
Therefore -h(x) =/= h(-x), making the equation NOT odd.

-g(x) = -(x^3-5x) = -(x^3)+5x
When solving for evens, we found that g(-x) = -(x^3)+5x, too.
Therefore, -g(x) = g(-x), making the equation odd.

That's explaining it algebraically, dude above gave you the answer in relation to how said equations appear graphically.


Hopefully I've explained that clearly.

Answer 3

f(x) is even y axis symmetric
h(x) neither
g(x) odd (0,0) symmetric

Answer 4

1st one is neither.
2nd one is odd.
3rd is neither.


i think.