How do you find the points of Inflection of #f(x)=2x(x-4)^3#?

2 Answers
Apr 4, 2018

Below

Explanation:

#f(x)=2x(x-4)^3#
#f'(x)=2xtimes3(x-4)^2+2(x-4)^3#
#f'(x)=2(x-4)^2(3x+x-4)=2(x-4)^2(4x-4)#
#f''(x)=2times(2(x-4)(4x-4)+(x-4)^2(4))#
#f''(x)=2(8x^2-40x+32+4x^2-32x+64)#
#f''(x)=2(12x^2-72x+96)#
#f''(x)=2(12)(x^2-6x+8)#
#f''(x)=24(x-4)(x-2)#

For points of inflexion, #f''(x)=0#

ie
#24(x-4)(x-2)=0#
#x=4,2#

Test #x=4#
At #x=3#, #f''(x)=-24#
At #x=4#, #f''(x)=0#
At #x=5#, #f''(x)=72#

Therefore, there is a change in concavity so there is a point of inflexion at x=4

Test #x=2#
At #x=1#, #f''(x)=72#
At #x=2#, #f''(x)=0#
At #x=3#, #f''(x)=-24#

Therefore there is a change in concavity so there is a point of inflexion at x=2

Apr 4, 2018

Point of inflection is at #x=4# or #x=8/3#

Explanation:

Points of Inflection appear where the curve changes from being concave to convex or vice versa. This happens when second derivative i.e. #(d^2f)/(dx^2)=0#

As #f(x)=2x(x-4)^3#

#(df)/(dx)=2(x-4)^3+2x(x-4)^2=2x^3-24x^2+96x-128+2x^3-16x^2+32x=4x^3-40x^2+128x-128#

and #(d^2f)/(dx^2)=12x^2-80x+128#

This is #0#, when #12x^2-80x+128=0#

or #3x^2-20x+32=0#

or #(x-4)(3x-8)=0#

Hence point of inflection is at #x=4# and #x=8/3#

Graph not drawn to scale.

graph{2x(x-4)^3 [-10, 10, -70, 30]}