How do you find intercepts, extrema, points of inflections, asymptotes and graph #y=x/(x^2+1)#?

1 Answer
Andrea S.
Jan 4, 2017

graph{x/(x^2+1) [-10, 10, -1, 1]}

Explanation:

The domain of the function is the entire #RR#, as the denominator of the rational function is always #>0#.

We have that:

#lim_(x->-oo) x/(x^2+1) = 0#

#lim_(x->+oo) x/(x^2+1) = 0#

We can see that #y(x)# has the line #y=0# as horizontal asymptote on both sides. We can also see that:

#y(x) < 0 # for #x<0#
#y(x) > 0 # for #x>0#
#y(x) = 0 # for #x=0#

So #x=0# is the only intercept.

#y'(x) = frac(1-x^2) ((x^2+1)^2)#

As the denominator of #y'(x)# is always positive the function is differentiable everywhere and:

#y'(x) <0# for #x in (-oo,-1)# and #in in (1,+oo)#
#y'(x) >0# for #x in (-1,1)#
#y'(x) = 0# for #x=+-1#

Therefore #y(x)# starts from #y=0# at #x->-oo# and decreases until #x=-1# where it reaches a local minimum #y(-1)=-1/2#. It then increases until #x=1# (changing sign at #x=0#) where it reaches a local maximum at #y(1) = 1/2#, and then decreases approaching zero indefinitely as #x->+oo#

#y''(x) = frac (2x(x^2-3)) ((x^2+1)^3)#

so inflection points are: #x=0# and #x=+-sqrt(3)# and the concavity is determined by the sign of #y''(x)#:

for #x in (-oo, -sqrt(3)), y''(x) <0, y(x)# is concave down
for #x in (-sqrt(3),0), y''(x) >0, y(x)# is concave up
for #x in (0, sqrt(3)), y''(x) <0, y(x)# is concave down
for #x in (sqrt(3),+oo), y''(x) >0, y(x)# is concave up

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