If #y = 1 / (1+x^2)#, what are the points of inflection, concavity and critical points?
1 Answer
First, let's note what each of these mean for a 2D graph on the Cartesian xyplane.
DEFINITIONS
A critical point for some function

#(df(c))/(dx) = 0# (an extremum, or perhaps an inflection point) 
#(df(c))/(dx)# does not exist (a corner, cusp, etc)
...where
Concavity is essentially a descriptor noting that on either side of a critical point, the graph's behavior is the same, i.e. the graph increases on both sides, or decreases on both sides.
They exist where the first derivative is
#0# but the second derivative is not#0# , but either positive or negative.
An inflection point is almost like a point of concavity, but on either side of the critical point, the behavior is opposite. Thus, we call inflection points the point where the concavity changes when moving across the specified critical point.
They exist when the first derivative is
#0# , with the condition that the second derivative is also#0# .
An example of inflection points? Try taking the second derivative of
THE GRAPH ITSELF
Before we start, yes, it kind of spoils where the critical point is, but here's the graph:
graph{1/(1+x^2) [10, 10, 5, 5]}
CRITICAL POINT
You can find a critical point by taking the first derivative. All you know from the critical point, however, is that the derivative is
For
#color(blue)((df(x))/(dx)) = (1 + x^2)^(2)*2x#
#= color(blue)((2x)/(1 + x^2)^2)#
If we set this to equal
#(df(c))/(dx) = 0 = (2x)/(1 + x^2)^2#
#= 2x#
#color(blue)(x = 0)#
Okay, so there is only one! Great. What is its concavity? Up or down?
CONCAVITY
To determine the concavity of the single inflection point, just take the second derivative of
#\mathbf((d^2f)/(dx^2) = (d[f"'"(x)])/(dx))#
#(d^2f)/(dx^2) = d/(dx)[(2x)/(1 + x^2)^2]#
Using the Quotient Rule and Chain Rule, we get:
#= ((1 + x^2)^2 (2)  (2x)(2(1 + x^2)*2x))/(1 + x^2)^4#
#= (2(1 + x^2)^(cancel(2)^(1)) + 8x^2cancel((1 + x^2)))/(1 + x^2)^(cancel(4)^(3))#
#= (2(1 + x^2) + 8x^2)/(1 + x^2)^3#
#= color(blue)((6x^2  2)/(1 + x^2)^3)#
Now, if we check the
#(d^2f(c))/(dx^2): #
#color(blue)((d^2f(0))/(dx^2)) = (6(0)^2  2)/(1 + (0)^2)^3#
#= color(blue)((2)/(1)^3 < 0)#
Since
INFLECTION POINTS
I'm going to say right upfrontthis graph has no inflection points. It only has one extremuma local maximum, which is concave down. It can't change concavity without becoming a new graph.
You can prove this by checking the second derivative at