Point of discontinuity

Point Discontinuity

What is a Rational Function?

Before getting into simplifying rational expressions and their restrictions, it is important to understand the rational function definition. Rational expressions are simply equations that feature a denominator that is also a function. They can be simple or widely complex. The presence of a function in the denominator makes rational expressions unique compared to other polynomials, as they also often feature what are called “restrictions” or restricted values. These values are restrictions on variables due to the fact that a denominator equal to zero is impossible – we cannot divide by zero because it is undefined. These restricted values can also be called discontinuities, and can present themselves either as asymptotes or as points – which we’ll talk more about later. Below are a few examples of rational expressions, as well as some simple restricted values of rational expressions.

x3x2(x0)\frac{x}{3x^{2}} (x \neq 0)

3x7(x0)\frac{3}{x^{7}} (x \neq 0)

4x2+2x2(x2)\frac{4x^{2}+2}{x-2} (x \neq 2)

Oftentimes, when dealing with rational expressions, we first need to conduct what is known as simplification. That is, trying to make the polynomial in the numerator and the denominator as simple as possible before we try to solve. For a look at how to simplify rational expressions, check out our video on rational expressions that focusses on this topic. As well, for more basic info on rational functions, check out our video on rational functions.

What is Discontinuity?

Discontinuity, in its most basic sense is exactly what you would suspect – an area or point(s) of a function that are discontinuous from the rest. Finding discontinuities is an important step in the process to solve rational functions. The most common examples of discontinuities are asymptotes (either a vertical asymptote or a horizontal asymptote) or a point discontinuity, otherwise referred to as a removable discontinuity. The term removable discontinuity comes from the fact that oftentimes the point discontinuity is a term that can be cancelled out from the numerator and denominator, making it “removable”.

The discontinuity definition with regards to points of discontinuity is simply a single point (with both an x and y-value) for which a function is undefined. This can be found in any function, not just rational ones.

The discontinuity definition with regards to asymptotic discontinuity, either horizontal or vertical, is a value for x (vertical) or y (horizontal) whereby the function rapidly approaches, but never touches. In other words, vertical asymptotes are values that make a function undefined in the domain of a rational function, and horizontal asymptotes are values that make a function undefined in the range of a rational function. This can be visualized better by looking at some of the examples later in this article, as well as the next couple of sections.

How to Find Horizontal Asymptote of a Rational Function:

In order to find a horizontal asymptote of a rational function, the process is actually quite simple, and can broken down into two methods:

i) Graphically:

\bullet Find value(s) of y that the function approaches exponentially but never touches

ii) Algebraically

\bullet If the degree of the denominator is larger than the degree of the numerator, the horizontal asymptote is the x-axis

\bulletIf the degree of the numerator is larger than the denominator, there is no horizontal asymptote

\bulletIf the degrees of the numerator and denominator are equal, the horizontal asymptote equals the leading coefficient (the coefficient value on the largest exponent) of the numerator divided by the leading coefficient of the denominator

How to Find Domain Asymptote of a Rational Function:

In order to find a vertical (domain) asymptote of a rational function, the process is actually quite simple, and can broken down into two methods:

i) Graphically:

\bullet Find value(s) of x that the function approaches exponentially but never touches

ii) Algebraically

\bullet Set the denominator equal to zero, and solve for x – these value(s) will be the vertical asymptote(s)

Now, let’s take a look at some examples of discontinuity to illustrate these points.

Example 1:

Sketch and compare the following two functions:

i) f(x)=2x+5f(x) = 2x+5

ii) g(x)=2x2+11x+15x+3g(x) = \frac{2x^{2}+11x+15}{x+3}

  • Notice if we factor g(x), we are able to “remove” (x+3) by cancelling it out. Also notice that x \neq -3 because otherwise the function would be undefined. This is the very definition of a “removable” or point discontinuity. Therefore, we can say we have a point discontinuity at (-3,-1).

    2x2+11x+15x+3=2x+5\frac{2x^{2}+11x+15}{x+3} = 2x+5

  • Otherwise, these two equations are exactly the same.
    Point discontinuity whether the equations are the same
    Point discontinuity whether the equations are the same
    Point discontinuity whether the equations are the same
    Point discontinuity whether the equations are the same

Example 2:

Sketch the rational function:

f(x)=2x27x+52x311x2+19x10f(x) = \frac{2x^{2}-7x+5}{2x^{3}-11x^{2}+19x-10}

  • First, let’s factor and simplify the rational function:

    (2x5)(x1)(2x5)(x1)(x2)=1x2\frac{(2x-5)(x-1)}{(2x-5)(x-1)(x-2)} = \frac{1}{x-2}

  • Using the same analysis as in Example 1, we can find several values for which this rational function is undefined:

    x52,1,2x \neq \frac{5}{2}, 1, 2

  • Furthermore, because we can remove several of these points, we have points of discontinuity at (52\frac{5}{2},2) and (1,-1). Because we can’t remove (x-2), but the function is still undefined at this x-value, we can say that x=2 is a vertical asymptote.

    Vertical asymptote where x is undefined at 2
    Vertical asymptote where x is undefined at 2

And that’s all there is to it! For further study, see our videos on quadratic functions, the limits at vertical asymptotes, and the limits at horizontal asymptotes.

Point of discontinuity

Lessons

Notes:
• "point of discontinuity" exists when the numerator and denominator have a factor in common.
i.e. (x)=(3x8)(x+5)(2x7)(x+5)(4x+9)(3x+8)(2x7)\left( x \right) = \frac{{ - \left( {3x - 8} \right)\left( {x + 5} \right)\left( {2x - 7} \right)}}{{\left( {x + 5} \right)\left( {4x + 9} \right)\left( {3x + 8} \right)\left( {2x - 7} \right)}} ; points of discontinuity exist at x=5x = - 5 and x=72x = \frac{7}{2} .
• To determine the coordinates of the point of discontinuity:
1) Factor both the numerator and denominator.
2) Simplify the rational expression by cancelling the common factors.
3) Substitute the non-permissible values of x into the simplified rational expression to obtain the corresponding values for the y-coordinate.
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Point of discontinuity

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