If #y = 1 / (1+x^2)#, what are the points of inflection, concavity and critical points?

A critical point for some function #f(x)# exists wherever some #f(c)# exists such that one of the following is true:

#(df(c))/(dx) = 0# (an extremum, or perhaps an inflection point)

#(df(c))/(dx)# does not exist (a corner, cusp, etc)

...where #c# takes on a chosen value within the domain of #f(x)#.

They exist where the first derivative is #0# but the second derivative is not #0#, but either positive or negative.

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 #x^3#; its critical point is at #x = 0# since #(d(x^3))/(dx) = 3x^2# and if #0 = 3x^2#, #x = 0#.

You can find a critical point by taking the first derivative. All you know from the critical point, however, is that the derivative is #0#. You do not know yet whether it is a maximum, minimum, or inflection point.

For #f(x) = 1/(1+x^2)#, using the Power Rule and the Chain Rule, the derivative is:

#color(blue)((df(x))/(dx)) = -(1 + x^2)^(-2)*2x#

#= color(blue)((-2x)/(1 + x^2)^2)#

If we set this to equal #0#, we can search for the value(s) of #x# that gives the critical point.

#(df(c))/(dx) = 0 = (-2x)/(1 + x^2)^2#

#= -2x#

#color(blue)(x = 0)#

To determine the concavity of the single inflection point, just take the second derivative of #f(x)#. Or, take the first derivative of the first derivative (#f'(x)#, or #(df)/(dx)#); same thing.

#\mathbf((d^2f)/(dx^2) = (d[f"'"(x)])/(dx))#

#(d^2f)/(dx^2) = d/(dx)[(-2x)/(1 + x^2)^2]#

#= ((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 #x# value where we found the critical point, we can look at the sign of this second derivative. If it's negative, the critical point is concave down, and vice versa.

#(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 #(d^2f(c))/(dx^2)# is negative, the graph is concave down at #x = 0#.

You can prove this by checking the second derivative at #x = c = 0#. If the second derivative had been #0# instead of negative or positive, then an inflection point would be present. But it's negative.

To find the points of inflection, concavity, and critical points of the function \( y = \frac{1}{1+x^2} \), we need to first find its second derivative, and then analyze its behavior. Given \( y = \frac{1}{1+x^2} \), the first derivative is: \[ \frac{dy}{dx} = -\frac{2x}{(1+x^2)^2} \] The second derivative is: \[ \frac{d^2y}{dx^2} = \frac{2(3x^2 - 1)}{(1+x^2)^3} \] To find critical points, solve for \( \frac{dy}{dx} = 0 \): \[ -\frac{2x}{(1+x^2)^2} = 0 \] \[ x = 0 \] To find concavity and points of inflection, set the second derivative equal to zero and solve: \[ \frac{d^2y}{dx^2} = 0 \] \[ 2(3x^2 - 1) = 0 \] \[ x^2 = \frac{1}{3} \] \[ x = \pm \frac{1}{\sqrt{3}} \] Now, we can analyze the concavity and determine points of inflection: When \( x < -\frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} > 0 \), so concave up. When \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} < 0 \), so concave down. When \( x > \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} > 0 \), so concave up. Thus, the points of inflection are \( x = -\frac{1}{\sqrt{3}} \) and \( x = \frac{1}{\sqrt{3}} \). The critical point is \( x = 0 \).