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
Counter Counter example (spoiler)
