How do you find points of inflection and determine the intervals of concavity given #y=4x^2e^(3x)#?

1 Answer
Sep 27, 2017

Use the second derivative #(d^2y)/(dx^2)# to find possible points of inflection, and then test those by evaluating the second derivative using test points in intervals around those points.

Explanation:

First find #dy/dx# using the Chain Rule:

#dy/dx = (4x^2)(3*e^(3x)) + (e^(3x))(2*4x)#

#=12x^2e^(3x) + 8xe^(3x)#
#=4xe^(3x)(3x + 2)#

Next, find #(d^2y)/(dx^2)# by differentiating again.

#(d^2y)/(dx^2) = (4xe^(3x))(3) + (3x+2)[(4x)(3*e^(3x))+(e^(3x)*4)]#
#=12xe^(3x)+(3x+2)(12xe^(3x)+4e^(3x))#
#=12xe^(3x)+36x^2e^(3x)+12xe^(3x)+24xe^(3x)+8e^(3x)#
#=36x^2e^(3x)+48xe^(3x)+8e^(3x)#
#=4e^(3x)(9x^2+12x+2)#

Potential points of inflection will occur at values of #x# where the second derivative is 0 or undefined. By examining #(d^2y)/(dx^2)# as derived, there are no places where it is undefined. To find places where it is 0, we set #(d^2y)/(dx^2) = 0# and solve:

#(d^2y)/(dx^2) = 0#
#4e^(3x)(9x^2+12x+2) = 0 => 4e^(3x) = 0 or (9x^2+12x+2) = 0#

#4e^(3x)=0# has no solutions.

#9x^2+12x+2 = 0# requires the Quadratic Formula:

#x=(-12+-sqrt(12^2-4*9*2))/(2*9)=(-12+-sqrt(144-72))/18#
#=(-12+-sqrt(72))/18=(-12+-6sqrt(2))/18=(-2+-sqrt(2))/3#

This results in values of #x~~-0.195,-1.1.38#

To determine if either are points of inflection, and to determine intervals of concavity, test values of #x# that "surround" these possible points of inflection, paying close attention to the sign of the second derivative at each test point. We can use #x# values of -2, -1, and 0. Note that #4e^(3x)# is always positive, and therefore the sign of the second derivative depends solely upon the second quadratic term:

#x=-2 => 9(-2)^2+12(-2)+2=36-24+2=62>0#
#x=-1 => 9(-1)^2+12(-1)+2=9-12+2=-1<0#
#x=0 => 9(0)^2+12(0)+2=0+0+2=2>0#

If the second derivative is positive, the function is concave up in that interval. Thus, in the intervals #(-oo, (-2-sqrt(2))/3)# and #((-2+sqrt(2))/3,oo)# the function is concave up, while in the interval #((-2-sqrt(2))/3,(-2+sqrt(2))/3)# the function is concave down.

Since each value for #x# determined above is a place where the concavity changes, each of those values of #x# are points of inflection. This can be verified by examining the graph of the function:

graph{4x^2e^(3x) [-2.496, 0.542, -0.557, 0.962]}