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

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Answer 1 · 2016-01-07T13:29:09

$x = 3$ $x=3$ vertical asymptote
$x = 7$ $x=7$ vertical asymptote

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

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

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

$x \in] - \infty, 3 [\cup] 3, 7 [\cup] 7, + \infty [$ $x in ]-oo,3[uu]3,7[uu]7,+oo[$

because

$f (x) = \frac{N (x)}{D (x)}$ $f(x)=(N(x))/(D(x))$

FE (Field of Existence):

$D (x) \neq 0$ $D(x)!=0$

$(x - 3) (x - 7) \neq 0$ $(x-3)(x-7)!=0$

$x_{1} \neq 3$ $x_1!=3$
$x_{2} \neq 7$ $x_2!=7$

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

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

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

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

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

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

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