Determining Points of Inflection for a Function
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M91: concavity and inflection points
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Calculus V.
Key Questions

#f(x)=x^3+x# By taking derivatives,
#f'(x)=3x^2+1# #f''(x)=6x=0 Rightarrow x=0# ,which is the
#x# coordinate of a possible inflection point. (We still need to verify that#f# changes its concavity there.)Use
#x=0# to split#(infty,\infty)# into#(infty,0)# and#(0,infty)# .Let us check the signs of
#f''# at sample points#x=1# and#x=1# for the intervals, respectively.
(You may use any number on those intervals as sample points.)#f''(1)=6<0 Rightarrow f# is concave downward on#(infty,0)# #f''(1)=6>0 Rightarrow f# is concave upward on#(0,infty)# Since the above indicates that
#f# changes its concavity at#x=0# ,#(0,f(0))=(0,0)# is an inflection point of#f# .I hope that this was helpful.

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Graphing with the Second Derivative

1Relationship between First and Second Derivatives of a Function

2Analyzing Concavity of a Function

3Notation for the Second Derivative

4Determining Points of Inflection for a Function

5First Derivative Test vs Second Derivative Test for Local Extrema

6The special case of x⁴

7Critical Points of Inflection

8Application of the Second Derivative (Acceleration)

9Examples of Curve Sketching