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

1 Answer
Jan 7, 2016

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