How do you determine where the graph of the given function is increasing, decreasing, concave up, and concave down for #h(x) = (x^2) / (x^2+1)#?

1 Answer
Hubert
May 28, 2015

First we need to evaluate the domain and the first and second derivatives:
#D_h=RR#
#h'(x)=(2x)/(x^2+1)^2#
#h''(x)=(-2(3x^2-1))/(x^2+1)^3=(-6(x+sqrt(3)/3)(x-sqrt(3)/3))/(x^2+1)^3#

Now, #h# is increasing when #h'>0# and decreasing when #h'<0#. Notice that #forall_(x in RR)\ x^2+1>0# so as long and we're interested only in the sing of the derivative we can ommit the denominator.
#2x>0 iff x>0# - function #h# increases when #x in (0;+oo)#
#2x<0 iff x<0# - function #h# decreases when #x in (-oo;0)#

Function #h# is concave up when #h''>0# and concave down when #h''<0#.
concave up: #-6(x+sqrt(3)/3)(x-sqrt(3)/3)>0 iff x in (-sqrt(3)/3;sqrt(3)/3)#
concave down: #-6(x+sqrt(3)/3)(x-sqrt(3)/3)<0 iff #
#iff x in (-oo;-sqrt(3)/3) cup (sqrt(3)/3;+oo)#

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