Determining Points of Inflection for a Function

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Concavity, Inflection Points and Second Derivatives
10:23 — by patrickJMT

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Key Questions

  • Make the second derivative equation equal to 0, and find the x values.

    When the second derivative is above 0, the original function is con caved up. If it is below 0, the original function is concave down.

  • #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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