Removing Discontinuities

Removing Discontinuities: AP Calculus AB/BC Study Guide

Hey mathlete! 🧮 Ready to tackle those pesky discontinuities in your graphs? Just think of discontinuities like potholes on your math highway. We're going to fill them in and make your journey as smooth as butter. Let's dive into the world of removing discontinuities and make your graphs continuous!

Discontinuities are those annoying breaks, jumps, or holes in your graph where the function just won't behave. There are three types: removable, jump, and infinite. Today, we're focusing on removable discontinuities because, unlike an unruly teenager, you can actually fix them! 💫

Imagine you're putting together a puzzle, and there's one piece missing - a glaring hole in your masterpiece. Removable discontinuities are just like that. They occur when a function isn't defined at a certain point, but the limit exists there. With a bit of math magic, we can insert that missing piece and make the function continuous.

Consider the function: [ f(x) = \frac{x - 1}{x^2 - 1} ]

This function has a removable discontinuity (a hole) at (x = 1). Why? Because when (x = 1), the denominator becomes zero, making the function undefined. But by factoring the denominator, we can fix it: [ f(x) = \frac{x - 1}{(x - 1)(x + 1)} = \frac{1}{x + 1}, \quad x \neq 1 ]

Now, instead of a hole, we're left with a continuous function everywhere, except we've filled in that discontinuity at (x = 1).

Let's try another example! Consider: [ g(x) = \frac{x^2 - 4x + 3}{x - 1} ]

First, factor the numerator: [ g(x) = \frac{(x - 1)(x - 3)}{x - 1} ]

Now, cancel out the common term: [ g(x) = x - 3, \quad x \neq 1 ]

To remove the discontinuity at (x = 1), set ( g(1) = 1 - 3 = -2 ). Voilà! Discontinuity be gone. ✨

Piecewise functions can be like trying to blend different genres in a playlist. To ensure your playlist, um, function is smooth (continuous), check the limits as (x) approaches the fun curve change point from the left and right, ensuring they match the defined function value.

Consider the piecewise function: [ f(x) = \begin{cases} x^2 - 3, & \text{if } x \leq 2 \ x - 1 + 2, & \text{if } x > 2 \end{cases} ]

For ( f ) to be continuous at ( x = 2 ):

• From the left: ( \lim_{x \to 2^-} f(x) = 2^2 - 3 = 1 )
• From the right: ( \lim_{x \to 2^+} f(x) = 2 - 1 + 2 = 1 )
• Function value at ( x = 2 ): ( f(2) = 1 )

Since all values agree, ( f(x) ) is as smooth as a baby’s bottom at ( x = 2 ).

Test out your skills! Try to make the function: [ h(x) = \begin{cases} \frac{x^2 + 5x + 4}{a(x + 4)}, & x \neq 2 \ a, & x = 2 \end{cases} ] continuous at ( x = 2 ).

Factor and simplify: [ \frac{x^2 + 5x + 4}{a(x + 4)} = \frac{(x + 1)(x + 4)}{a(x + 4)} = \frac{x + 1}{a}, \quad x \neq 2 ]

At ( x = 2 ): [ \frac{2 + 1}{a} = \frac{3}{a} = a \implies a^2 = 3 \implies a = \sqrt{3} ]

So, for ( h ) to be continuous at ( x = 2 ), ( a ) must be ( \sqrt{3} ).

Use graphing tools to visualize continuity. A continuous graph can be drawn without lifting your pencil, just like a good story can be told without interruptions. ✍️ Using a simple graphing calculator or software, plot your functions and look for those dreaded discontinuities. If they exist, put your newfound skills to the test and try to "patch them up."

Wrapping Up

Congratulations! You've navigated the treacherous path of discontinuities and emerged a calculus conqueror. Practice consistently with problems from your textbook and past AP exams to strengthen your understanding. Always check for continuity and practice removing those mathematical "potholes". Happy calculating! 🎉