#f(x)={x^2}/{x^2+3}#
By Quotient Rule,
#f'(x)={2x cdot(x^2+3)-x^2 cdot 2x}/{(x^2+3)^2}={6x}/{(x^2+3)^2}=0#
#=>x=0# (the only critical number)
By checking the sample values #x=pm1#,
#{(f'(-1)>0),(f'(1)<0):}=>{(f'(x) < 0 " on "(-infty,0)),(f'(x)>0" on "(-infty,0)):}#,
Hence,
#f# is #{("increasing on "[0,infty)),("decreasing on "(-infty,0]):}#
By First Derivative Test,
#f':(-) to (+)# aound #x=0# #=> f(0)=0# is a local minimum.
(Note: There is no local maximum.)
By Quotient Rule,
#f''(x)=6{1cdot(x^2+3)^2-x cdot2(x^2+3)(2x)}/{(x^2+3)^4}={18(1+x)(1-x)}/{(x^2+3)^3}=0#
#=> x=pm1#
By checking the sample values #x=-2,0,2#,
#{(f''(-2)<0),(f''(0)>0),(f''(2)<0):} =>
{(f''(x)<0" on "(-infty,-1)),(f''(x)>0" on "(-1,1)),(f''(x)<0" on "(1,infty)):}#
Hence,
#f# is #{("concave upward on "(-1,1)),("concave downward on "(-infty,-1)cup(1,infty)):}#
Since the concavity changes at #x=pm1#, the #x#-ccordinates of the inflection points are #x=pm1#.
I hope that this was helpful.
