How to Find Discontinuities of Rational Functions?

To find the points of discontinuity of a rational function, set the denominator equal to zero and solve.

Related Topics

• How to Find the End Behavior of Rational Functions
• How to Graph Rational Functions

A step-by-step guide to the discontinuities of rational functions

In rational functions, points of discontinuity refer to fractions that are undefined or have zero denominators. When the denominator of a fraction is (0), it becomes undefined and appears as a hole or a break in the graph.

To find discontinuities of rational functions, follow these steps:

• Obtain a function’s equation. If the numerator and denominator share a factor, cancel it, but its zero is still a discontinuity: a hole in the graph.
• Rewrite the denominator expression as a zero-valued equation.
• Solve the equation for the denominator.

The Discontinuities of Rational Functions, Example 1:

Find the discontinuities of (f(x)=frac{x-1}{x^2-x-6}).

First, setting the denominator equal to zero: (x^2-x-6=0).

Then factoring it out: (x^2-x-6=0) ⇒ ((x+2)(x-3)=0)

(x+2=0 ⇒ x=-2)

(x-3=0 ⇒x=3)

Now, (f) is discontinuous at (x=-2) and (x=3).

The Discontinuities of Rational Functions, Example 2:

Find the discontinuities of (f(x)=frac{1}{x^2-4}).

First, setting the denominator equal to zero: (x^2-4=0).

Then factoring it out: (x^2-4=0) ⇒ ((x+2)(x-2)).

(x+2=0) ⇒ (x=-2)

(x-2=0) ⇒ (x=2)

Now, (f) is discontinuous at (x=-2) and (x=2).

Exercises for the Discontinuities of Rational Functions

Find the discontinuities of rational functions.

1. (color{blue}{f(x)=frac{x+2}{x^2-5x-6}})
2. (color{blue}{f(x)=frac{x-2}{x^2-2x-35}})
3. (color{blue}{f(x)=frac{x^2-6x+8}{x-5}})
4. (color{blue}{f(x)=frac{x+10}{x^2-10x+21}})

1. (color{blue}{x=-1, x=6})
2. (color{blue}{x=7, x=-5})
3. (color{blue}{x=5})
4. (color{blue}{x=3, x=7})

Holes vs. Vertical Asymptotes

Every zero of the denominator is a point of discontinuity, but not every one looks the same on a graph. The difference comes from whether the factor cancels.

Worked Example: Finding a Hole and an Asymptote

Find all points of discontinuity of f(x) = (x² − 9)/(x² − x − 6).

1. Factor both parts. x² − 9 = (x − 3)(x + 3) and x² − x − 6 = (x − 3)(x + 2).
2. Find where the denominator is 0. (x − 3)(x + 2) = 0 gives x = 3 and x = −2. The function is discontinuous at both.
3. Cancel the common factor. (x − 3) cancels, leaving (x + 3)/(x + 2) for x ≠ 3.
4. Classify. The canceled factor (x − 3) gives a hole at x = 3. The factor (x + 2) is still in the denominator, so x = −2 is a vertical asymptote.
5. Find the hole’s y-value. Plug x = 3 into the simplified form: (3 + 3)/(3 + 2) = 6/5. The hole is at (3, 6/5).

A common mistake is plugging x = 3 into the original function. That gives 0/0, which tells you nothing. Always use the simplified expression for the y-value of a hole.

Frequently Asked Questions

What is a point of discontinuity in a rational function?

It is an x-value where the function is undefined, so the graph has a break. For a rational function, these points are exactly the x-values that make the denominator equal 0. Each one is either a hole, where a factor cancels, or a vertical asymptote, where a factor stays in the denominator.

How do you tell a hole from a vertical asymptote?

Factor the numerator and denominator completely. If a factor appears in both and cancels, its zero is a hole. If a factor is left in the denominator after canceling, its zero is a vertical asymptote. For example, (x + 2)/((x + 2)(x − 5)) has a hole at x = −2 and an asymptote at x = 5.

How do you find the coordinates of a hole?

Set the canceled factor equal to 0 to get the x-value. Then substitute that x-value into the simplified function, not the original one. For f(x) = (x² − 4)/(x − 2), the simplified form is x + 2, so the hole is at x = 2 and y = 4, or the point (2, 4).

Can a rational function have no points of discontinuity?

Yes. If the denominator is never 0 for any real number, the function is continuous everywhere. For example, f(x) = 1/(x² + 1) has no discontinuities, because x² + 1 is always at least 1. Solving x² + 1 = 0 gives no real solutions.

Is a removable discontinuity the same as a hole?

Yes. A removable discontinuity is the formal name for a hole. It is called removable because you could fill in that single missing point and make the function continuous there. A vertical asymptote cannot be fixed that way, so it is called an infinite or non-removable discontinuity.