How do you find all points of inflection given #y=x^2/(2x+2)#?

Question

Luke Alvarez · Accepted Answer

For #y=x^2/(2x+2)#, there are no inflection points, but there is a vertical asymptote at #x=-1.#

Inflection points are the points where the graph's concavity changes—that is, where the slopes of the graph go from increasing to decreasing (or vice versa). Visually, inflection points are the points where the graph curves like a bowl on one side (concave up) and like a hill on the other (concave down). (e.g. #f(x)=sinx# has inflection points at every point where the graph crosses the #x#-axis—at #x=0,+-pi,+-2pi,#etc.)

Mathematically, inflection points occur where the second derivative is equal to 0. So we need to find #y''#.

For #y=x^2/(2x+2)#, we use the division rule to get

#y'=((2x)(2x+2)-x^2(2))/(2x+2)^2#

#color(white)(y')=(4x^2+4x-2x^2)/[2(x+1)]^2#

#color(white)(y')=(2(x^2+2x))/(4(x+1)^2)#

#color(white)(y')=(x^2+2x)/(2(x+1)^2)#

(We can leave the numerator un-factored to help us find its derivative again.)

Now, we take this #y'# and find its derivative: #y''#.

#y''=((2x+2)(2)(x+1)^2-(x^2+2x)(2)(2)(x+1)(1))/[2(x+1)^2]^2#

#color(white)(y'')=(cancel4(x+1)^(cancel(3)2)-cancel4x(x+2)cancel((x+1)))/(cancel4(x+1)^(cancel(4)3))#

#color(white)(y'')=((x+1)^2-x(x+2))/((x+1)^3)#

#color(white)(y'')=(cancel(x^2)+cancel(2x)+1-cancel(x^2)-cancel(2x))/((x+1)^3)#

#color(white)(y'')=1/(x+1)^3#

We now set this second derivative equal to 0 and try to solve for #x#...

#1/(x+1)^3=0#

Uh-oh! We can't do it. #1/"something"# can never equal 0, no matter what that something is.

Because of this, there are no inflection points, but there will be a vertical asymptote when #2x+2=0# (in other words, when #x=-1.)#

Christopher Agresta · Answer

undefined To find all points of inflection for the function $ y = \frac{x^2}{2x + 2} $, follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $ to find the points where the concavity may change.
3. Test the concavity of the function around those points to confirm whether they are points of inflection.

First, find the second derivative of $ y $ with respect to $ x $:

$$ y = \frac{x^2}{2x + 2} $$

$$ y' = \frac{d}{dx} \left( \frac{x^2}{2x + 2} \right) $$

Use the quotient rule:

$$ y' = \frac{(2x + 2)(2x) - (x^2)(2)}{(2x + 2)^2} $$

$$ y' = \frac{4x^2 + 4x - 2x^2}{(2x + 2)^2} $$

$$ y' = \frac{2x^2 + 4x}{(2x + 2)^2} $$

Now, find the second derivative:

$$ y'' = \frac{d}{dx} \left( \frac{2x^2 + 4x}{(2x + 2)^2} \right) $$

$$ y'' = \frac{(2x + 2)^2(4x) - (2x^2 + 4x)(2)(2x + 2)(2)}{(2x + 2)^4} $$

$$ y'' = \frac{8x(2x + 2)^2 - 8x(2x + 2)^2}{(2x + 2)^4} $$

$$ y'' = 0 $$

This equation has no solutions other than $ x = -1 $, where the denominator is zero.

Now, test the concavity around $ x = -1 $ to determine if it's a point of inflection:

- Choose test points on either side of $ x = -1 $, such as $ x = -2 $ and $ x = 0 $.
- Plug these test points into the second derivative.
- If the sign changes, then $ x = -1 $ is a point of inflection.

For $ x = -2 $:
$$ y''(-2) = \frac{8(-2)(2(-2) + 2)^2 - 8(-2)(2(-2) + 2)^2}{(2(-2) + 2)^4} $$
$$ y''(-2) = \frac{8(-2)(0) - 8(-2)(0)}{0} $$
$$ y''(-2) = 0 $$

For $ x = 0 $:
$$ y''(0) = \frac{8(0)(2(0) + 2)^2 - 8(0)(2(0) + 2)^2}{(2(0) + 2)^4} $$
$$ y''(0) = \frac{0 - 0}{2^4} $$
$$ y''(0) = 0 $$

Since the second derivative is zero at $ x = -1 $ and does not change sign, there are no points of inflection for the function $ y = \frac{x^2}{2x + 2} $.