$F(x,t)=a_n(t)x^n+ \ldots +a_1(t)x+a_0(t)$ is a through $t$ parametrized family of polynominals. $a_i : I \to \Bbb R \:\:\:\mathrm{ are }\: \mathcal C^k$- functions with $k \ge 1$. Let $x_0$ be a zero of degree $1$, of $F( \cdot , t_0) $ for $ t_0 \in I$, i.e.$$ F(x_0, t_0) =0 \:\:\:\:\: \mathrm{and} \:\:\:\:\: \frac{\partial F}{\partial x} (x_0, t_0 ) \not = 0$$ Using the Implicit Function Theorem i now have to show the following:
There exist neighborhoods $U$ of $x_0$ and $V$ of $t_0$ so that $F(\cdot , t)$ has exactly one zero in $U$, which depends $\mathcal C^k$-differentiable on $t$.
I know how to use the Implicit Function Theorem for regular Systems of Equations but I'm kind of lost here on how it can come out that there is only one exact zero.
Any ideas or tipps? Thanks in advance !
I think what confuses here is that we want to express $x$ in terms of $t$ and not the other way round, i.e. we apply the implicit function theorem to the first variable. Now, since $\frac{\partial F}{\partial x}(x_0, t_0) \neq 0$, the theorem is applicable, and you get neighborhoods $U$ of $x_0$ and $V$ of $t_0$ and a $C^k$-function $g: V \rightarrow U$ with $x_0 = g(t_0)$, such that
\begin{equation} F(x,t) = 0 \Leftrightarrow x = g(t) \Leftrightarrow F(g(t), t) = 0 \end{equation}
for all $(x,t) \in U \times V$. Now, since $g$ is a function it takes only one value for each $t$, and so your function $F(\cdot, t)$ which has it's $t$ fixed has exactly one zero in $U$, namely $g(t)$.