A correct statement of the tangent line test for inflection points Bourchtein and Bourchtein (2015) give a counterexample showing that the following statement of the tangent line test is not generally true:

if the tangent line at the point $c$ lies above (below) the graph of $f(x)$ in a left-hand neighborhood of $c$ and below (above) the graph of $f(x)$ in a right-hand neighborhood of $c$, then $c$ is an inflection point.

What assumptions would we have to add so that the above statement becomes true? For example, would it suffice if $f$ is continuously differentiable (on a neighborhood around $c$)? Twice differentiable?

Their counterexample:



Their definitions:

 A: Define
$$f(x)=\begin{cases} e^{-1/x}(1.1+\sin(1/x)),& x >0 \\  0 ,& x=0 \\-e^{-1/x}(1.1+\sin(1/x)), & x <0        \\
 \end{cases}$$
Then $f\in C^\infty(\mathbb R),$ with $D^nf(0)=0$ for all $n.$ The tangent line at $(0,0)$ is thus the line $y=0.$ We have $f>0$ for $x>0,$ $f<0$ for $x<0.$
For $x>0$ we have
$$f'(x)= e^{-1/x}x^{-2}(1.1+\sin(1/x)-\cos(1/x)).$$
Thus
$$f'(x)>0 \text { for  } x=\frac{1}{3\pi/4+2n\pi}, n = 1,2,\dots$$
and
$$f'(x)<0 \text { for  } x=\frac{1}{-\pi/4+2n\pi}, n = 1,2,\dots$$ 
It follows that $f$ is neither convex or concave in every interval $(0,r).$ The same is true to the left of $0.$ Therefore $0$ is not an inflection point of $f$ (according to the definition you gave).
But suppose $f$ equals a power series in a neighborhood of $c.$ Then the proposed result is valid. You might want to try this, so I'll leave it to you for now. Ask if you have questions.

Added later
Assume that $g(x) = \sum_{n=2}^{\infty }a_nx^n$ in $(-r,r), $ and $g>0$ on $(0,r),$ $g<0$ on $(-r,r).$ Then $0$ is an inflection point of $g.$ (This implies the desired result.)
Proof: Let $m$ be the first integer such that $a_m\ne 0.$ We then have $g(x) = a_mx^m + O(x^{m+1})$ as $x\to 0.$ For small $x$ the $a_mx^m$ term dominates. Since $g>0$ in $(0,r),$ we must have $a_m>0.$ If $m$ were even, then $g<0$ in $(-r,0)$ would fail. Thus $m$ is odd. Now 
$$g''(x) = m(m-1)a_mx^{m-2} + O(x^{m-1}).$$
It follows that $g''(x)>0$ for small $x>0.$ Furthermore, since $m-2$ is also odd, $m(m-1)a_mx^{m-2}<0$ for small $x<0.$ This implies $g''(x)<0$ for small $x<0.$ Therefore $0$ is an inflection point of $g,$ and we're done.
