How do you find the inflection points of the graph of the function: #f(x) = 2x(x-4)^3#?

1 Answer
Jun 6, 2015

The inflection points are given when the second derivative is equaled to zero, because the second derivative indicates the change in concavity of the function.

#(dy)/(dx)=2(x-4)^3+2x(3(x-4)^2)=2(x-4)^3+6x(x-4)^2#

#(dy)/(dx)=[(2x-8)+(6x)][(x-4)^2]=color(green)((8x-8)(x-4)^2)#

#(d^2y)/(dx^2)=8(x-4)^2+(8x-8)(2x-8)#

Developing:

#(d^2y)/(dx^2)=8(x^2-8x+16)+(16x^2-80x+64)=24x^2-144x+192#

Equaling the second derivative to zero:

#24(x^2-6x+8)=0#

#(6+-sqrt(36-(4)(1)(8)))/2#
#(6+-sqrt(4))/2#
#(6+-2)/2#

#x_1=4# and #x_2=2#

Substituting these values found in the second derivative there in the original function, we'll get the inflection points.

#f(4)=2(4)(4-4)^3=0#
#f(2)=2(2)(2-4)^3=4*(-8)=-32#

Thus, the inflection points are #A(4,0)# and #B(2,-32)#