For what values of x, if any, does #f(x) = 1/((x-3)(x-7)) # have vertical asymptotes?

1 Answer
Jan 7, 2016

Answer:

#x=3# vertical asymptote
#x=7# vertical asymptote

graph{1/((x-3)(x-7)) [-10, 10, -5, 5]}

Explanation:

Vertical asymptotes could be find in the points where the Exist Conditions of #f(x)# are not satisfied

The Field of Existence of #f(x)# is:

#x in ]-oo,3[uu]3,7[uu]7,+oo[#

because

#f(x)=(N(x))/(D(x))#

FE (Field of Existence):

#D(x)!=0#

#(x-3)(x-7)!=0#

#x_1!=3#
#x_2!=7#

Now these values of #x# are vertical asymptotes if:

#lim_(x rarr x_(1,2)^+-)f(x)=+-oo#

#lim_(x rarr 3^-)f(x)=lim_(x rarr 3^-)1/((2.9-3)(2.9-7))=#
#lim_(x rarr 3^-)1/((0^-)(-4.1))=lim_(x rarr 3^-)1/0^+=+oo#

#lim_(x rarr 3^+)f(x)=lim_(x rarr 3^+)1/((3.1-3)(3.1-7))=#
#lim_(x rarr 3^+)1/((0^+)(-3.9))=lim_(x rarr 3^+) 1/(0^-) =-oo#

#lim_(x rarr 7^-)f(x)=lim_(x rarr 7^-)1/((6.9-3)(6.9-7))=#
#lim_(x rarr 7^-)1/((3.9)(0^-))=lim_(x rarr 7^-)1/(0^-)=-oo#

#lim_(x rarr 7^+)f(x)=lim_(x rarr 7^+)1/((7.1-3)(7.1-7))=#
#lim_(x rarr 7^+)1/((4.1)(0^+))=lim_(x rarr 7^+)1/(0^+)=+oo#