P is a polynomial of degree 6. M and N are two real numbers with 0 < M < N.

Given that:

- P(M) = P(-M), and:
- P(N) = P(-N)
- P’(0) = 0

Does the relationship P(x) = P(-x) hold for all nonzero real values of x?

If so, prove it.

If not, provide a counterexample.

Yes, it is true, and it holds for all values of x (zero or non-zero).

We only need to consider the odd powered terms.

Let Q(x) = bx^5 + dx^3 + fx

From 3, we know that f = 0

So Q(x) = bx^5 + dx^3

We have bM^5 + dM^3 = b(-M)^5 + d(-M)^3

If b <> 0 then d/b = -M^2

Similarly,

If b <> 0 then d/b = -N^2

This is a contradiction, so b = 0, and so d = 0.

So P(x) = ax^6 + cx^4 + ex^2 + g

So P(x) = -P(x) for all x