How do you sketch the graph of f(x)=(x+1)^(-1)???

1 Answer
May 18, 2017

This is equivalent to saying

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

This is a rational function that will have vertical asymptotes at #x= -1#. Now we must derive the equation of the horizontal asymptote.

#y = lim_(x->oo) (1/x)/(x/x + 1/x)#

#y = lim_(x->oo) (1/x)/(1 + 1/x)#

#y = (lim_(x->oo) 1/x)/(lim_(x->oo) 1 + lim_(x->oo) 1/x)#

#y= 0/(1 + 0)#

#y= 0#

Therefore, there will be a horizontal asymptote at #y= 0#. Now find the invariant points. These will occur at #y = +-1#.

#1 = 1/(x +1)#

#x + 1 = 1#

#x= 0#

Hence, #(0, 1)# will also serve as a y-intercept.

AND

#-1 = 1/(x + 1)#

#-1(x + 1) = 1#

#-x - 1 = 1#

#-x = 2#

#x= -2#

So, the graph resembles the following:

graph{y = 1/(x + 1) [-10, 10, -5, 5]}

Hopefully this helps!