How do you find the inflection point of this graph? The graph is this: 
$$
\frac{(x+1)^3 - 4(x+1)^2 + 4(x+1)}{(x+1)^2 - 2(x+1) + 1}
$$
I know you can find second derivative and then solve for values that make it undefined or 0, but I was told apparently there is another faster way to get the inflection point. What is that method?
 A: simplifying the given term we get $$\frac{(x-1)^2(x+1)}{x^2}$$
i think yes since the second and the third derivative is simple
A: Let $$y=\frac{(x+1)^3 - 4(x+1)^2 + 4(x+1)}{(x+1)^2 - 2(x+1) + 1}$$
Then using the substitution, $u=x+1$, we obtain
$$y=\frac{u^3 - 4u^2 + 4u}{u^2 - 2u + 1}$$
This simplifies to $$y=\frac{u(u-2)^2}{(u-1)^2}$$
Differentiating with respect to $u$, we obtain
$$\frac{\partial y}{\partial u}=\frac{u^3-3u^2+4u-4}{(u-1)^3}$$
Since $$\frac{\partial u}{\partial x}=1$$
We have $$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\cdot\frac{\partial u}{\partial x}=\frac{\partial y}{\partial u}$$
Substituting back in $u=x+1$, we are left with
$$\frac{\partial y}{\partial x}=\frac{(x+1)^3-3(x+1)^2+4(x+1)-4}{x^3}$$
This simplifies to 
$$\frac{\partial y}{\partial x}=\frac{(x^3+3x^2+3x+1)-(3x^2+6x+3)+(4x+4)-4}{x^3}=\frac{x^3+x-2}{x^3}$$
Computing the second derivative
$$\frac{\partial^2 y}{\partial x^2}=\frac{\partial}{\partial x}(1+\frac{1}{x^2}-\frac{2}{x^3})=-\frac{2}{x^3}+\frac{6}{x^4}=\frac{6-2x}{x^4}$$
A point of inflection will occur when $\frac{\partial y}{\partial x}=0$, solving for this we get $6-2x=0$ gives $x=3$
The sufficient condition for points of inflection require that either side of the neighbourhood of $x=3$ have different signs.
Checking in the neighbourhood of $x=3$ we obtain a positive value for $\frac{\partial^2 y}{\partial x^2}$ when $x=2$ and a negative value for $\frac{\partial^2 y}{\partial x^2}$ when $x=4$.
Therefore $x=3$ is a point of inflection.
