Regarding point of inflection Statement $1$: At point of inflection $\dfrac{dy}{dx}=0$ or doesn't exist.
Statement $2$: If a function is strictly increasing or strictly decreasing at some point and  $\dfrac{dy}{dx}=0$ at that point, then that point is point of inflection.
Comment on the correctness of these statements, give examples to support your claim.
My attempt is as follows:-
Statement $1$: At point of inflection $\dfrac{d^2y}{dx^2}=0$ or doesn't exist. But we can't say anything about $\dfrac{dy}{dx}$. 
$$f(x)=x^3-3x^2+2x-1$$
$$f'(x)=3x^2-6x+2$$
$$f''(x)=6x-6$$
$f''(x)=0$ at $x=1$ and concavity of $f(x)$ changes at $x=1$ but $f'(1)=3+2-6=-1$
Statement $2$:
Suppose if the function is strictly increasing and $\dfrac{dy}{dx}=0$ at $x=x_0$ then $f(x_0-h)<f(x_0)<f(x_0+h)$ and $\dfrac{dy}{dx}>0$ in the neighborhood.
But from this information I am not getting how to prove that $x_0$ is the point of inflection.
 A: Consider
$$g(x)=\begin{cases}|x\sin\frac1x|&x\ne 0\\0&x=0\end{cases} $$
and let $f(x)=\int_0^xg(t)\,\mathrm dt$.


*

*As $g$ is continuous, we have $g(x)=f'(x)$. In particular, $f'(0)=0$.

*As $g$ is non-negative, $f$ is non-decreasing.

*In fact, whenever $a<b$, there exists an open interval  $(u,v)\subset [a,b]$ where $g$ is strictly positive; therefore $f$ is strictly increasing.

*For all $\epsilon>0$, the interval $(0,\epsilon)$ contains intervals where $g'(x)$ (exists and) is positive and others where $g'(x)$ is negative. Hence $f$ is neither convex nor concave on $(0,\epsilon)$; the same holds for $(-\epsilon,0)$.


We conclude that $0$ is not a point of inflection for $f$.
A: Let $f$ be a function differentiable at some interior point $x_0$ of its domain, let there exist a $\delta_0 > 0$ such that $f$ is strictly increasing on $\left( x_0 - \delta_0, x_0 + \delta_0 \right)$, and let $f^\prime \left( x_0 \right) = 0$.
The equation of the tangent to the curve $y = f(x)$ at the point $\left( x_0, f \left( x_0 \right) \right)$ is given by
$$ y - f\left( x_0 \right) = f^\prime \left( x_0 \right) \left( x- x_0 \right), $$
which reduces to
$$ y = f\left( x_0 \right). $$
Thus the curve $y = f(x)$ has a horizontal tangent at the point $\left( x_0, f \left( x_0 \right) \right)$.
As $f(x) < f\left(x_0 \right)$ for any point $x \in \left( x_0 - \delta_0, x_0 \right)$, so immediately to the left of the vertical line $x = x_0$ our curve is below the tangent line.
And, as $f(x) > f\left(x_0 \right)$ for any point $x \in \left( x_0, x_0 + \delta_0 \right)$, so immediately to the right of the vertical line $x = x_0$, our curve is above the tangent line.
Thus our curve crosses its tangent line at the point $\left( x_0, f \left( x_0 \right) \right)$.
