Piecewise Functions - ACT Math

1. What does each sub-function in a piecewise function have?

   Its own domain, specified by an interval or condition. Each piece is restricted to specific input values.

2. What is the correct way to solve $f(x)=k$ for a piecewise function?

   Solve $\text{branch}(x)=k$ and keep only solutions in that branch interval. Check solutions fall within each piece's domain.

3. Find $k$ so $f$ is continuous at $x=0$: $f(x)=\begin{cases}k-x,&x<0\\x+2,&x\ge^0\end{cases}$.

   $k=2$. Set left limit $k = 2$ to match right value.

4. Find $f(3)$ for $f(x) = \begin{cases} x + 3 & \text{if } x < 3 \\ 5x - 1 & \text{if } x \geq 3 \end{cases}$.

   $14$. Since $3 \geq 3$, use $5x - 1$: $5(3) - 1 = 14$.

5. Identify the sub-function for $x < -1$ in $f(x) = \begin{cases} 3x + 2 & \text{if } x < -1 \\ x^2 + 1 & \text{if } x \geq -1 \end{cases}$.

   $3x + 2$. This is the first sub-function for the given condition.

6. Evaluate $f(-1)$ for $f(x)=\begin{cases}x^2,&x\le^0\\2x+1,&x>0\end{cases}$.

   $1$. Since $-1 \le 0$, use $x^2 = (-1)^2 = 1$.

7. Identify whether $f$ is continuous at $x=1$: $f(x)=\begin{cases} x+4, & x<1 \\ x^2, & x \geq 1 \end{cases}$.

   Not continuous at $x=1$. Left limit $5$ doesn't equal right value $1$.

8. How do you identify sub-functions in a piecewise function?

   Each sub-function is defined by a specific interval or condition. Each piece has its own domain restriction.

9. What is the first step to evaluate a piecewise function at $x=a$?

   Select the branch whose condition includes $x=a$. Check which interval contains the input value.

10. Identify the sub-function at $x = 4$ for $f(x) = \begin{cases} x^3 & \text{if } x < 4 \\ 2x + 5 & \text{if } x \geq 4 \end{cases}$.

    $2x + 5$. Since $4 \geq 4$, this sub-function applies.

11. Identify which inequality symbol, $<$ or $\le$, includes the endpoint in an interval.

    $\le$ includes the endpoint. $<$ excludes the endpoint value.

12. What is the piecewise definition of $|x-a|$?

    $|x-a|=\begin{cases}x-a,&x\ge a\\a-x,&x<a\end{cases}$. Expression inside absolute value changes sign at $x=a$.

13. What is the key feature of a piecewise function?

    It changes its rule or formula based on the input value. The function definition varies based on input conditions.

14. Solve $f(x)=4$ for $f(x)=\begin{cases}x+2,&x<1\\2x,&x\ge^1\end{cases}$.

    $x=2$. Since $x \ge 1$, solve $2x = 4$ giving $x = 2$.

15. Solve $f(x)=2$ for $f(x)=\begin{cases}x^2,&x<0\\x+2,&x\ge^0\end{cases}$.

    $x=-\sqrt{2}$. Since $x < 0$, solve $x^2 = 2$ giving $x = -\sqrt{2}$.

16. Evaluate $|{-3}|$ using the piecewise definition of $|x|$.

    $3$. Since $-3 < 0$, use $-x = -(-3) = 3$.

17. What is the first step in evaluating a piecewise function?

    Determine which sub-function applies to the given input value. Check which condition the input satisfies first.

18. Find $f(2)$ for $f(x) = \begin{cases} 5 & \text{if } x < 1 \\ x^2 & \text{if } x \geq 1 \end{cases}$.

    $4$. Since $2 \geq 1$, use $x^2$: $2^2 = 4$.

19. What point style on a graph represents an included endpoint, $\le$ or $\ge$?

    A closed (filled) dot. Filled dots show the point is part of the function.

20. Find $c$ so $f$ is continuous at $x=1$: $f(x)=\begin{cases}3x+c,&x<1\\x^2,&x\ge^1\end{cases}$.

    $c=-2$. Set left limit $3 + c = 1$ to match right value.

21. Identify the sub-function for $x \geq 0$ in $f(x) = \begin{cases} x^3 & \text{if } x < 0 \\ 4x & \text{if } x \geq 0 \end{cases}$.

    $4x$. This is the second sub-function for non-negative inputs.

22. Identify the interval notation for the condition $-2<x\le^5$.

    $(-2,5]$. Parenthesis excludes $-2$, bracket includes $5$.

23. Determine $f(-2)$ for $f(x) = \begin{cases} x^2 + 4 & \text{if } x < 0 \\ x + 5 & \text{if } x \geq 0 \end{cases}$.

    $8$. Since $-2 < 0$, use $x^2 + 4$: $(-2)^2 + 4 = 8$.

24. Identify the correct endpoint inclusion for $x\in[1,3)$ in inequality form.

    $1\le x<3$. Square bracket includes, parenthesis excludes endpoints.

25. Identify $f(0)$ for $f(x)=\begin{cases}1,&x<0\\2,&x>0\end{cases}$.

    $f(0)$ is undefined. No piece includes $x = 0$ in its domain.

26. What is the purpose of using piecewise functions?

    To model situations where a rule or relationship changes based on different conditions. Handles cases where different rules apply to different inputs.

27. Find $a$ so the pieces match at $x=3$: $f(x)=\begin{cases}x+a,&x<3\\2x-1,&x\ge^3\end{cases}$.

    $a=2$. Set left limit $3 + a = 5$ to match right value.

28. Identify the sub-function applied at $x = 0$ in $f(x) = \begin{cases} -x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$.

    $x^2$. Since $0 \geq 0$, the second sub-function applies.

29. Identify the sub-function for $x < 2$ in $f(x) = \begin{cases} x^3 - x & \text{if } x < 2 \\ 2x^2 & \text{if } x \geq 2 \end{cases}$.

    $x^3 - x$. This is the first sub-function for the given condition.

30. Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.

    $5$. Since $1 \geq 1$, use $2x^2 + 3$: $2(1)^2 + 3 = 5$.

31. What is an essential property of a piecewise function's graph?

    It may have different slopes or curvatures in different intervals. Each piece can have distinct mathematical behavior.

32. Find $f(0)$ for $f(x) = \begin{cases} x^2 - 3 & \text{if } x < 0 \\ 3x + 4 & \text{if } x \geq 0 \end{cases}$.

    $4$. Since $0 \geq 0$, use $3x + 4$: $3(0) + 4 = 4$.

33. Evaluate $f(3)$ for $f(x)=\begin{cases}4,&x<1\\x-2,&x\ge^1\end{cases}$.

    $1$. Since $3 \ge 1$, use $x - 2 = 3 - 2 = 1$.

34. Rewrite $f(x)=|x|+1$ as a piecewise function.

    $f(x)=\begin{cases}x+1,&x\ge^0\\-x+1,&x<0\end{cases}$. Split $|x|$ at zero and add 1 to each piece.

35. What is the absolute value identity as a piecewise definition for $|x|$?

    $|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$. Definition splits at zero where sign changes.

36. Evaluate $f(5)$ for $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$.

    $10$. Apply the third piece: $2(5) = 10$.

37. Find $f(0)$ for $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 3x + 2 & \text{if } x \geq 0 \end{cases}$.

    $2$. Since $0 \geq 0$, use $3x + 2$: $3(0) + 2 = 2$.

38. Find $b$ so $f$ is continuous at $x=-1$: $f(x)=\begin{cases}x^2+b,&x\le-1\\2x,&x>-1\end{cases}$.

    $b=-3$. Set left value $1 + b = -2$ to match right limit.

39. What is a piecewise function?

    A function defined by multiple sub-functions, each applying to a certain interval. Different rules apply to different input ranges.

40. What is an example of a piecewise function?

    $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 3 & \text{if } x \geq 0 \end{cases}$. Shows quadratic behavior for negatives, linear for non-negatives.

41. Which branch gives $f(5)$ if $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$?

    Use $2x$ because $x\ge^4$. Check which condition contains $x = 5$.

42. What is the notation for a piecewise function?

    Uses curly braces and specifies sub-functions with different conditions. Mathematical notation showing multiple conditional cases.

43. Determine $f(1)$ for $f(x) = \begin{cases} 7 - x & \text{if } x < 1 \\ 2x & \text{if } x \geq 1 \end{cases}$.

    $2$. Since $1 \geq 1$, use $2x$: $2(1) = 2$.

44. Identify the range of $f(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ 2x - 3 & \text{if } x \geq 0 \end{cases}$.

    All real numbers. Both linear pieces cover all possible output values.

45. What is the range of a piecewise function?

    The set of all possible output values of the function. Combination of all outputs from each sub-function.

46. Evaluate $f(2)$ for $f(x)=\begin{cases}x+1,&x<2\\3x,&x\ge^2\end{cases}$.

    $6$. Since $2 \ge 2$, use $3x = 3(2) = 6$.

47. Identify whether $f$ is continuous at $x=1$: $f(x)=\begin{cases}x+4,&x<1\\x^2,&x\ge^1\end{cases}$.

    Not continuous at $x=1$. Left limit $5$ doesn't equal right value $1$.

48. Determine $f(-4)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ -x + 2 & \text{if } x \geq 0 \end{cases}$.

    $16$. Since $-4 < 0$, use $x^2$: $(-4)^2 = 16$.

49. What is a characteristic of the transition between sub-functions in a piecewise function?

    The transition may be continuous or discontinuous. Depends on whether boundary values align between pieces.

50. Determine $f(-3)$ for $f(x) = \begin{cases} 2x + 1 & \text{if } x \leq 0 \\ x^2 - 4 & \text{if } x > 0 \end{cases}$.

    $-5$. Since $-3 \leq 0$, use $2x + 1$: $2(-3) + 1 = -5$.

51. What is the key challenge in graphing piecewise functions?

    Ensuring correct transitions and connections between sub-functions. Must handle discontinuities and different function behaviors.

52. Find $f(4)$ for $f(x) = \begin{cases} 2x - 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}$.

    $14$. Since $4 \geq 3$, use $x^2 - 2$: $4^2 - 2 = 14$.

53. Evaluate $f(0)$ for $f(x)=\begin{cases}5-x,&x<0\\x^2+2,&x\ge^0\end{cases}$.

    $2$. Since $0 \ge 0$, use $x^2 + 2 = 0 + 2 = 2$.

54. Find $f(-1)$ for $f(x) = \begin{cases} x^3 + 1 & \text{if } x \leq 0 \\ 4x^2 & \text{if } x > 0 \end{cases}$.

    $0$. Since $-1 \leq 0$, use $x^3 + 1$: $(-1)^3 + 1 = 0$.

55. Identify the sub-function for $x \geq 3$ in $f(x) = \begin{cases} x + 2 & \text{if } x < 3 \\ 4x & \text{if } x \geq 3 \end{cases}$.

    $4x$. This is the second sub-function in the definition.

56. Find the value at the boundary: $f(1)$ for $f(x)=\begin{cases}x+4,&x<1\\x^2,&x\ge^1\end{cases}$.

    $1$. Since $1 \ge 1$, use second piece: $1^2 = 1$.

57. Find $f(-5)$ for $f(x) = \begin{cases} x - 3 & \text{if } x < -2 \\ x^2 & \text{if } x \geq -2 \end{cases}$.

    $-8$. Since $-5 < -2$, use $x - 3$: $-5 - 3 = -8$.

58. What is the domain of a piecewise function?

    The set of all real numbers that the function is defined for. Union of all intervals where sub-functions are defined.

59. Find $a$ so $f$ is continuous at $x=2$: $f(x)=\begin{cases}ax+1,&x<2\7,&x \geq 2\end{cases}$.

    $a=3$. Set left limit $2a + 1 = 7$ to match right value.

60. Evaluate $|2-7|$ using the piecewise definition of $|x-a|$.

    $5$. Since $2 - 7 = -5 < 0$, use $-(-5) = 5$.

61. Determine $f(0)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases}$.

    $1$. Since $0 \geq 0$, use $x + 1$: $0 + 1 = 1$.

62. Find $a$ so $f$ is continuous at $x=2$: $f(x)=\begin{cases}ax+1,&x<2\\7,&x\ge^2\end{cases}$.

    $a=3$. Set left limit $2a + 1 = 7$ to match right value.

63. What is the meaning of $f(c)$ when the piecewise definition includes $x=c$ in one branch?

    $f(c)$ equals the formula value from the branch with condition including $x=c$. Use the formula from the piece containing that $x$-value.

64. What must be true for a piecewise function to be continuous?

    Sub-functions must connect at their boundaries without jumps. Function values must match at boundary points.

65. Rewrite $f(x)=|x-2|$ as a piecewise function.

    $f(x)=\begin{cases}x-2,&x\ge^2\\2-x,&x<2\end{cases}$. Split where $x - 2$ changes sign at $x = 2$.

66. What is the correct way to solve $f(x)=g(x)$ when $f$ is piecewise?

    Set each branch equal to $g(x)$ and restrict solutions to that branch interval. Apply each piece separately and verify domain restrictions.

67. What is a common application of piecewise functions?

    Modeling tax rates, shipping costs, or situations with different conditions. Real-world scenarios with varying rules or rates.

68. Solve $f(x)=0$ for $f(x)=\begin{cases}x+3,&x\le-1\\x-1,&x>-1\end{cases}$.

    $x=-3$ and $x=1$. Solve in both pieces: $x + 3 = 0$ and $x - 1 = 0$.

69. Identify the sub-function for $x < 0$ in $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 3x + 1 & \text{if } x \geq 0 \end{cases}$.

    $x^2$. Since $-3 < 0$, use the first sub-function.

70. Determine $f(5)$ for $f(x) = \begin{cases} 3x - 1 & \text{if } x < 2 \\ x^2 + 2 & \text{if } x \geq 2 \end{cases}$.

    $27$. Since $5 \geq 2$, use $x^2 + 2$: $5^2 + 2 = 27$.

71. Evaluate $f(-2)$ for $f(x)=\begin{cases}2x+3,&x\le-1\\x^2,&x>-1\end{cases}$.

    $-1$. Since $-2 \le -1$, use $2x + 3 = 2(-2) + 3 = -1$.

72. Solve $f(x)=1$ for $f(x)=\begin{cases}2-x,&x<1\\x^2,&x\ge^1\end{cases}$.

    $x=1$. Since $x \ge 1$, solve $x^2 = 1$ giving $x = 1$.

73. What point style on a graph represents an excluded endpoint, $<$ or $>$?

    An open (hollow) dot. Hollow dots show the point is not included.

74. Determine $f(6)$ for $f(x) = \begin{cases} 2x - 5 & \text{if } x < 4 \\ x^2 + 1 & \text{if } x \geq 4 \end{cases}$.

    $37$. Since $6 \geq 4$, use $x^2 + 1$: $6^2 + 1 = 37$.

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