Polynomial Graphs in the neighbourhood of zero

Question

Given a polynomial $P(x) = a_1 x + a_2 x^2 + .. + a_n x^n$ with $a_0 = 0$

I am trying to prove that in the neighbourhood of zero $B_{r}(0)$ the graph of the polynomial will cross the x-axis if and only if the smallest integer $k \in \{1,...,n\}$ such that $a_{k} \neq 0$ is odd

I know that an odd-degree polynomial with $P(0) = 0$ will cross the x-axis while an even-degree polynomial wont, but I am confused with degree of the smallest term, for instance $x^4 + x^5$ has an odd degree so it crosses the x-axis in general, but in the neighbourhood of zero it won't cross, because the smallest $k$ is even (the odd power is also high, so it's flattened out at zero)

Any help would be great!

Answer 1

Suppose $k$ is odd and $P(x)=a_kx^k+\cdots+a_nx^n=x^k(a_k+\cdots a_nx^{n-k})=x^kg(x)$. As $x\to 0$, $g(x)\to a_k$. Assume $a_k>0$, otherwise consider $-P$ so in a small enough neighborhood of zero, $0<a_k-\epsilon\le g(x)$. If $x>0$, $0<(a_k-\epsilon)x^k\le P(x)$. If $x<0$, $0<>a_k-\epsilon)x^k\ge P(x)$. IVT yields the result.

Answer 2

We have

$P(x) = \displaystyle \sum_k^n a_ix^i = a_k x^k \sum_k^n \dfrac{a_i}{a_k} x^{i - k} = a_k x^k \left (1 + \dfrac{a_{k + 1}}{a_k}x + \dfrac{a_{k + 1}}{a_k}x^2 + \ldots \dfrac{a_n}{a_k} x^{n - k} \right ); \tag 1$

we note the polynomial

$Q(x) = 1 + \dfrac{a_{k + 1}}{a_k}x + \dfrac{a_{k + 1}}{a_k}x^2 + \ldots \dfrac{a_n}{a_k} x^{n - k} \tag 2$

is a continuous function of $x$ such that

$Q(0) = 1; \tag 3$

it follows that there is a $\delta > 0$ such that for $\vert x \vert < \delta$ , the quantity

$Q(x) > 0, \tag 4$

so for $0 \ne \vert x \vert < \delta$, the sign of

$P(x) = a_k x^k Q(x) \tag 5$

is the same as that of $a_k x^k$, which changes at $0$ if $k$ is odd, but not if $k$ is even.