Determining Points of Inflection for a Function

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.

  • No. Consider #f(x)=x# - this function's concavity does not change throughout the entire run of the function.

    All polynomials with odd degree of 3 or higher have points of inflection, and some polynomials of even degree (again, higher than 3) have them. The best way to determine if a function has a point of inflection is to look at its second derivative - if the second derivative can equal zero, the original function has a point of inflection.

Questions