Analyse the function:

#f(x) = (x^2+1)/x#

A. This is a rational function so it is defined and continuous for all real numbers except the roots of the denominator, that is in #RR-{0}#

We can note that:

#f(-x) = ((-x)^2+1)/(-x) = -(x^2+1)/x = -f(x)#

so the function is odd.

B. The numerator #x^2+1# is always positive and #> 1#, so the function has no roots and #f(x) > 0# for #x>0# and #f(x) < 0# for #x <0#. This means the graph of the function has no intercepts.

C. For #x->0# we have:

#lim_(x->0^-) f(x) = -oo#

#lim_(x->0^+) f(x) = oo#

so the line #y=0# is a vertical asymptote.

Futhermore:

#lim_(x->-oo) f(x) =-oo#

#lim_(x->+oo) f(x) =oo#

so the the function does not have horizontal asymptotes. However:

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

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

so the line #y=x# is an asympote for #x->+-oo#

D. Evaluate the first derivative:

#(df)/dx = d/dx ((x^2+1)/x) = d/dx (x+1/x) =1-1/x^2#

thus the critical points for which #(df)/dx = 0# are in #x=+-1# and we can see that:

#(df)/dx > 0 # for #abs x > 1# and #(df)/dx < 0 # for #abs x < 1#

Then #f(x)# is monotone increasing in #(-oo,-1)# and #(1,oo)# while it is monotone decreasing in #(-1,0)# and #(0,1)#. It follows that in #x=-1# the function has a local maximum with value #f(-1) = -2# and in #x=1# it has a local minimum with value #f(1) = 2#.

E. Evaluate the second derivative:

#(d^2f)/dx^2 = d^2/dx^2( (x^2+1)/x) = d/dx (1-1/x^2) = 2/x^3#

We have that #(d^2f)/dx^2 != 0# for every #x# so the function has no inflection points. Besides:

#(d^2f)/dx^2 < 0# for #x < 0# and #(d^2f)/dx^2 > 0# for #x > 0#

so the function is concave down in #(-oo,0) # and concave up in #(0,oo)#.

graph{ (x^2+1)/x [-20, 20, -10, 10]}