How do you find points of inflection and determine the intervals of concavity given #y=x^3-3x^2+4#?
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#2|k^2# then #2|k# for some #k\inZZ#"
1 Answer
Nov 28, 2016
concave down on #(-infinity,1) and concave up on (1, infinity)
x=1 is an inflection point.
Explanation:
# y'=3x^2-6x# #y'' = 6x-6 or 6(x-1)# - prepare sign chart with x=1 as a critical point using second derivative.
- to the left of x=1, y'' is negative so concave down. to right of x=1, it is positive so concave up.
- since the concavity changes at x=1, it is an inflection point.
