#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.