Let $f∈C^{(n+1)}(a,b)$and suppose $x_o∈(a,b)$ and $f′(x_o) =···=f^{(n)}(x_0) = 0$ but $f^{(n+1)}(x_o) \neq 0$. Let $f∈C^{(n+1)}(a,b)$and suppose $x_o∈(a,b)$ and $f′(x_o) =···=f^{(n)}(x_0) = 0$ but $f^{(n+1)}(x_o) \neq 0$.  Then in $x_o$ the function f has
(i)  a strict local minimum, if $n$ is odd and $f^{(n+1)}(x_o)>0$,
(ii)  a strict local maximum, if $n$ is odd and $f^{(n+1)}(x_o) < 0$
(iii)  no extremum, if $n$ is even.
My work
Not a lot actually. I definitely do understand why the properties are as such. But where I am having trouble is explicitly writing it down in appropriate mathematical language. As such, what would be a "model answer" for questions like this that prove the statements conclusively?

Here are some big hints.
Since $f^{(n+1)}(x_0)\neq 0$ and $f^{(n+1)}$ is continuous, there exists $\delta$ such that $f^{(n+1)}(x)$ has the same sign as $f^{(n+1)}(x_0)$ for all $x$ with $|x-x_0|<\delta$.
Then apply Taylor's theorem. For all $x$ with $|x-x_0|<\delta$, we may write
$$
f(x)=f(x_0)+f'(x_0)(x-x_0)+\ldots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}
$$
for some $c$ between $x$ and $x_0$. It should be obvious how to break it up into cases and determine when $f(x)>f(x_0)$ and when $f(x)<f(x_0)$.
