Removing Discontinuities - AP Calculus AB

1. How do you simplify $f(x) = \frac{x^2 - 1}{x - 1}$?

   $x + 1$, after factoring and canceling $(x-1)$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1$.

2. Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x = 3$.

   No, it has a removable discontinuity at $x = 3$. The function is undefined at $x=3$ but has a finite limit there.

3. What is the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$?

   At $x = 5$, as $f(x)$ can be redefined as $x + 5$. Factor $x^2-25=(x-5)(x+5)$ to cancel $(x-5)$ and get $x+5$.

4. What does it mean if $\lim_{x \to a} f(x)$ does not exist?

   The function has a non-removable discontinuity at $x = a$. Either one-sided limits differ or approach infinity.

5. How can you remove a discontinuity in $f(x) = \frac{x^2 - 9}{x - 3}$?

   Redefine $f(x)$ at $x = 3$ as $f(x) = x + 3$. Factor $x^2-9=(x-3)(x+3)$ to cancel $(x-3)$ and get $x+3$.

6. What indicates a removable discontinuity in a function's graph?

   A hole in the graph where a point can be filled. Removable discontinuities appear as holes that can be filled.

7. What type of discontinuity does a piecewise function often have?

   Jump discontinuity, if the pieces do not connect. Different function rules at boundaries often create jumps.

8. How do you redefine $f(x) = \frac{x^2 - 16}{x - 4}$ to remove the discontinuity?

   Redefine $f(x)$ at $x = 4$ as $f(x) = x + 4$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.

9. What is the removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$?

   At $x = 1$, as $f(x)$ can be redefined as $x + 1$. Factor $x^2 - 1 = (x-1)(x+1)$ to cancel $(x-1)$ and get $x+1$.

10. Identify the removable discontinuity in $f(x) = \frac{x^2 - 9}{x + 3}$.

    At $x = -3$, as $f(x)$ can be redefined by simplifying. Factor $x^2-9=(x+3)(x-3)$ to cancel $(x+3)$ and get $x-3$.

11. What is the removable discontinuity in $h(x) = \frac{x^2 - 1}{x + 1}$?

    At $x = -1$, as $h(x)$ can be redefined by simplifying. Factor $x^2-1=(x+1)(x-1)$ to cancel $(x+1)$ and get $x-1$.

12. Identify the type of discontinuity in $g(x) = \frac{1}{x - 3}$ at $x = 3$.

    Infinite discontinuity at $x = 3$. The denominator approaches zero causing infinite behavior.

13. When is a discontinuity non-removable?

    If $\nexists \, \lim_{x \to a} f(x)$; jump or infinite discontinuities. When limits don't exist or are infinite, discontinuities cannot be removed.

14. Identify the type of discontinuity in $f(x) = \frac{1}{x - 2}$ at $x = 2$.

    Infinite discontinuity at $x = 2$. The denominator approaches zero causing the function to approach infinity.

15. Is $f(x) = x^2 - 4$ continuous for all real numbers?

    Yes, it is a polynomial with no discontinuities. Polynomial functions have no breaks, holes, or asymptotes.

16. Can all discontinuities be removed?

    No, only removable discontinuities can be redefined to be continuous. Only holes can be filled; jumps and infinite breaks cannot.

17. Which type of discontinuity is present in $f(x) = \frac{1}{x}$ at $x = 0$?

    Infinite discontinuity at $x = 0$. The denominator approaches zero while numerator stays constant.

18. What is a common factor indicating a removable discontinuity?

    A factor that cancels in both numerator and denominator. Canceling common factors reveals removable discontinuities.

19. How can you remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$?

    Redefine $f(x)$ at $x = 2$ as $f(x) = x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and get $x+2$.

20. Describe a jump discontinuity.

    A discontinuity where the left and right limits exist but are unequal. The one-sided limits differ, creating a break in the function.

21. Find the removable discontinuity in $g(x) = \frac{x^2 - 1}{x - 1}$.

    At $x = 1$, as $g(x)$ can be redefined as $x + 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and simplify to $x+1$.

22. What is a removable discontinuity?

    A point where a function is not defined but can be redefined to make it continuous. This describes a hole that can be filled by redefining the function value.

23. For $f(x) = x^2 - 4$, where is the removable discontinuity?

    No removable discontinuity; $f(x)$ is continuous. Polynomial functions are continuous everywhere on their domains.

24. Identify the removable discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.

    At $x = 2$, as $f(x)$ can be redefined by simplifying to $x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and simplify to $x+2$.

25. How do you verify a removable discontinuity exists at $x = a$?

    Check if $\lim_{x \to a} f(x)$ exists and $f(a)$ is undefined or $\neq \lim_{x \to a} f(x)$. These conditions define exactly when a discontinuity can be removed.

26. To remove a discontinuity, what must be true about $\lim_{x \to a} f(x)$?

    It must exist and equal a finite value. Infinite limits cannot be used to remove discontinuities.

27. Which discontinuity type cannot be removed by redefining $f(x)$?

    Jump and infinite discontinuities. These involve limits that don't exist or are infinite.

28. Is the function $f(x) = \frac{x^2 - 1}{x - 1}$ continuous at $x = 1$?

    No, it has a removable discontinuity at $x = 1$. The function is undefined at $x=1$ but has a finite limit there.

29. What must be true for a function to be continuous at $x = a$?

    $f(a) = \lim_{x \to a} f(x)$ and both must exist. All three conditions ensure no gaps, jumps, or holes in the function.

30. What is the simplified form of $f(x) = \frac{x^2 - 16}{x - 4}$?

    $x + 4$, after factoring and canceling $(x-4)$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.

31. Which discontinuity is not found in $y = x^3 - 3x + 2$?

    No discontinuities; continuous for all real $x$. Polynomial functions are continuous everywhere with no discontinuities.

32. State the condition for a removable discontinuity at $x = a$.

    $\lim_{x \to a} f(x)$ exists, but $f(a)$ is not defined or $f(a) \neq \lim_{x \to a} f(x)$. The limit must exist but the function value is either undefined or different.

33. How do you remove a discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$?

    Redefine $f(x)$ as $f(x) = x + 1$ at $x = 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1=2$.

34. Which step can help identify a removable discontinuity in a rational function?

    Factor the numerator and denominator to find common factors. Common factors that cancel create removable discontinuities at their zeros.

35. What is the simplified form of $h(x) = \frac{x^2 + x - 6}{x - 2}$?

    $x + 3$, after factoring numerator and canceling $(x-2)$. Factor $x^2+x-6=(x-2)(x+3)$ to cancel $(x-2)$ and get $x+3$.

36. Does $f(x) = \frac{x^2 - 4x + 4}{x - 2}$ have a removable discontinuity?

    Yes, at $x = 2$, as $f(x)$ simplifies to $x - 2$. Factor $(x-2)^2$ in numerator to cancel $(x-2)$ leaving $x-2$.

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