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

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Answer 1 · 2018-07-02T15:07:50

#x=2#

Given function,

#f(x)=(x+2)(x-4)^2#

#f(x)=x^3-6x^2+32#

#f'(x)=3x^2-12x#

#f''(x)=6x-12#

For, inflection point #f''(x)=0#

# 6x-12=0#

#x=2#