Pre-Calculus - Find a Point of Discontinuity

Find the point of discontinuity for the following function:

$g(x)=\frac{x^2-1}{x^2+8x+7}$

$\left(-1, -\frac{1}{3}\right)$

There is no point of discontinuity.

$\left(1, \frac{1}{3}\right)$

Explanation

Start by factoring the numerator and denominator of the function.

$g(x)=\frac{x^2-1}{x^2+8x+7}=\frac{(x-1)(x+1)}{(x+1)(x+7)}=\frac{x-1}{x+7}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{-1-1}{-1+7}=-\frac{1}{3}$

$\left(-1, -\frac{1}{3}\right)$ is the point of discontinuity.

If possible, find the type of discontinuity, if any:

$y= \frac{6x-9}{2x-3}$

$\textup{Hole at }x=\frac{3}{2}$

$\textup{Asymptote at }x=\frac{3}{2}$

$\textup{Hole and asymptote at }x=\frac{3}{2}$

$\textup{Hole at }x=-\frac{3}{2}$

$\textup{There are no discontinuities.}$

Explanation

By looking at the denominator of $y= \frac{6x-9}{2x-3}$ , there will be a discontinuity.

Since the denominator cannot be zero, set the denominator not equal to zero and solve the value of .

$x eq\frac{3}{2}$

There is a discontinuity at $x=\frac{3}{2}$ .

To determine what type of discontinuity, check if there is a common factor in the numerator and denominator of $y= \frac{6x-9}{2x-3}$ .

Since the common factor is existent, reduce the function.

$y= \frac{6x-9}{2x-3}=\frac{(3)(2x-3)}{2x-3}=3$

Since the term can be cancelled, there is a removable discontinuity, or a hole, at $x=\frac{3}{2}$ .

Find the point of discontinuity in the function $f(x)=\frac{4x^3-1}{x-4}$ .

$x=\frac{1}{4}$

Explanation

When dealing with a rational expression, the point of discontinuity occurs when the denominator would equal 0. In this case, so . Therefore, your point of discontinuity is .

Find the point of discontinuity for the following function:

$h(x)=\frac{x^2-5x+6}{x^2-9}$

$\left(3, \frac{1}{6}\right)$

$\left(5, \frac{1}{8}\right)$

There is no point of discontinuity for this function.

$\left(6, -\frac{2}{3}\right)$

Explanation

Start by factoring the numerator and denominator of the function.

$h(x)=\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{3-2}{3+3}=\frac{1}{6}$

$\left(3, \frac{1}{6}\right)$ is the point of discontinuity.

Find the point of discontinuity for the following function:

$k(x)=\frac{2x^2-x-1}{x^2-6x+5}$

$\left(1, -\frac{3}{4}\right)$

There is no point of discontinuity for this function.

$\left(-5, \frac{9}{10}\right)$

Explanation

Start by factoring the numerator and denominator of the function.

$k(x)=\frac{2x^2-x-1}{x^2-6x+5}=\frac{(2x+1)(x-1)}{(x-1)(x-5)}=\frac{2x+1}{x-5}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{2(1)+1}{1-5}=-\frac{3}{4}$

$\left(1, -\frac{3}{4}\right)$ is the point of discontinuity.

Find the point of discontinuity for the following function:

$l(x)=\frac{2x^2+5x-12}{x^2+3x-4}$

$\left(-4, \frac{11}{5}\right)$

There is no point of discontinuity for this function.

Explanation

Start by factoring the numerator and denominator of the function.

$l(x)=\frac{2x^2+5x-12}{x^2+3x-4}=\frac{(2x-3)(x+4)}{(x+4)(x-1)}=\frac{2x-3}{x-1}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{2(-4)-3}{-4-1}=\frac{11}{5}$

$\left(-4, \frac{11}{5}\right)$ is the point of discontinuity.

Find the point of discontinuity for the following function:

$n(x)=\frac{x^3+2x^2-3x}{x^2+8x+15}$

$\left(-1, \frac{1}{2}\right)$

There is no point of discontinuity for this function.

Explanation

Start by factoring the numerator and denominator of the function.

$n(x)=\frac{x^3+2x^2-3x}{x^2+8x+15}=\frac{x(x^2+2x-3)}{(x+3)(x+5)}=\frac{x(x-1)(x+3)}{(x+3)(x+5)}=\frac{x(x-1)}{x+5}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{-3(-3-1)}{-3+5}=6$

is the point of discontinuity.

Find the point of discontinuity for the following function:

$m(x)=\frac{x^2+4x-5}{x^2+7x+10}$

There is no point of discontinuity for the function.

Explanation

Start by factoring the numerator and denominator of the function.

$m(x)=\frac{x^2+4x-5}{x^2+7x+10}=\frac{(x+5)(x-1)}{(x+5)(x+2)}=\frac{x-1}{x+2}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{-5-1}{-5+2}=2$

is the point of discontinuity.

For the following function, $y=\frac{x^2-6x+9}{x^2-9}$ , find all discontinuities, if possible.

$\textup{Asymptote at: } x=-3 \textup{; Hole at: x=3}$

$\textup{Asymptote at: } x=3 \textup{; Hole at: x=-3}$

$\textup{Hole at: }x=\pm3$

$\textup{Asymptote at: }x=\pm3$

$\textup{There are no discontinuities.}$

Explanation

Rewrite the function $y=\frac{x^2-6x+9}{x^2-9}$ in its factored form.

$y=\frac{x^2-6x+9}{x^2-9}=\frac{(x-3)(x-3)}{(x+3)(x-3)}$

Since the term can be cancelled, there is a removable discontinuity, or a hole, at .

The remaining denominator of indicates a vertical asymptote at .

Find the point of discontinuity for the following function:

$p(x)=\frac{x^3-4x^2-5x}{x^2-9x+20}$

There is no point fo discontinuity for this function.

Explanation

Start by factoring the numerator and denominator of the function.

$p(x)=\frac{x^3-4x^2-5x}{x^2-9x+20}=\frac{x(x^2-4x-5)}{(x-5)(x-4)}=\frac{x(x-5)(x+1)}{(x-5)(x-4)}=\frac{x(x+1)}{x-4}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.