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?

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    $\begingroup$ To begin with maybe put $x+1=u$ and then work with the simpler function of $u$ (remembering to "shift back" for the true $x$ coordinates). $\endgroup$ – coffeemath Apr 4 '15 at 16:42
  • $\begingroup$ Maybe you can simplify that expression before you try to take the second derivative. $\endgroup$ – Mike Pierce Apr 4 '15 at 16:42
  • $\begingroup$ I don't think there is a faster way to get a point of inflection other than finding the second derivative and its solution and checking for a sign change in its neighbourhood. $\endgroup$ – Sam Houston Apr 4 '15 at 18:01

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.

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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

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    $\begingroup$ Does this make finding inflection points easier? (If so how?) $\endgroup$ – coffeemath Apr 4 '15 at 16:50
  • $\begingroup$ @coffeemath To elaborate a little on how the doctor got this, notice that you can factor $x+1$ out of the numerator getting $(x+1)( [x+1]^2-4[x+1]+4)$ and the right-hand factor you can recognize as quadratic in $x+1$ which can be factored as $[x+1]^2 -4[x+1] + 4 = ([x+1]-2)^2$. If this isn't easy to see, follow the comment above and switch to $x+1=u$ and then try doing this, and at the end switch back to $x+1$. $\endgroup$ – Addem Apr 4 '15 at 16:54
  • $\begingroup$ @coffeemath Also, now that the expression is simplified, taking derivatives is easier. So ultimately you still compute two derivatives and set equal to $0$, etc. $\endgroup$ – Addem Apr 4 '15 at 16:55
  • $\begingroup$ @Addem Yes. So it still seems to remain tthat two derivatives are needed to be sure of inflections. $\endgroup$ – coffeemath Apr 4 '15 at 18:12
  • $\begingroup$ On the aftermath of seeing the techniques/solutions offered here, those mainly being either simplify the expression before differentiation, or substitute (x+1)=u, simplifying, then solving, I think it is pretty safe to assume that you must get the second derivative to solve it, but the way in which you get the second derivative can be made easier, which is probably what my friend meant, when he said "there is a faster way." Thanks for the help! I also feel dumb for not having thought of simplifying it or subbing x+1 as another variable. Guess I didn't put enough thought into it. Thanks again! $\endgroup$ – inglavatin Apr 5 '15 at 1:11

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