ACT Math Flashcards: Piecewise Functions

What this deck covers

This deck focuses on Piecewise Functions, giving you a quick way to review the definitions, rules, and examples that matter most for ACT Math.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

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

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

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

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

Flashcard 3: What is the correct way to solve $f(x)=k$ for a piecewise function?

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

Flashcard 4: Find $k$ so $f$ is continuous at $x=0$: $f(x)=\begin{cases}k-x,&x<0\\x+2,&x\ge^0\end{cases}$.

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

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

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

Flashcard 6: What point style on a graph represents an included endpoint, $\le$ or $\ge$?

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

Flashcard 7: Identify the sub-function for $x < -1$ in $f(x) = \begin{cases} 3x + 2 & \text{if } x < -1 \\ x^2 + 1 & \text{if } x \geq -1 \end{cases}$.

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

Flashcard 8: Evaluate $f(-1)$ for $f(x)=\begin{cases}x^2,&x\le^0\\2x+1,&x>0\end{cases}$.

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

Flashcard 9: Evaluate $f(-2)$ for $f(x)=\begin{cases}2x+3,&x\le-1\\x^2,&x>-1\end{cases}$.

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

Flashcard 10: Identify whether $f$ is continuous at $x=1$: $f(x)=\begin{cases} x+4, & x<1 \\ x^2, & x \geq 1 \end{cases}$.

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

Flashcard 11: What is the correct way to solve $f(x)=g(x)$ when $f$ is piecewise?

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

Flashcard 12: How do you identify sub-functions in a piecewise function?

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

Flashcard 13: Find $f(-5)$ for $f(x) = \begin{cases} x - 3 & \text{if } x < -2 \\ x^2 & \text{if } x \geq -2 \end{cases}$.

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

Flashcard 14: Find $b$ so $f$ is continuous at $x=-1$: $f(x)=\begin{cases}x^2+b,&x\le-1\\2x,&x>-1\end{cases}$.

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

Flashcard 15: Find $f(0)$ for $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 3x + 2 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 16: What is the first step to evaluate a piecewise function at $x=a$?

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

Flashcard 17: Identify the sub-function at $x = 4$ for $f(x) = \begin{cases} x^3 & \text{if } x < 4 \\ 2x + 5 & \text{if } x \geq 4 \end{cases}$.

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

Flashcard 18: Determine $f(-4)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ -x + 2 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 19: What is the first step to evaluate a piecewise function at $x=a$?

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

Flashcard 20: Identify the correct endpoint inclusion for $x\in[1,3)$ in inequality form.

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

Flashcard 21: Identify which inequality symbol, $<$ or $\le$, includes the endpoint in an interval.

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

Flashcard 22: What is the piecewise definition of $|x-a|$?

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

Flashcard 23: Identify the sub-function applied at $x = 0$ in $f(x) = \begin{cases} -x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 24: What is the key feature of a piecewise function?

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

Flashcard 25: Solve $f(x)=4$ for $f(x)=\begin{cases}x+2,&x<1\\2x,&x\ge^1\end{cases}$.

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

Flashcard 26: Identify the interval notation for the condition $-2<x\le^5$.

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

Flashcard 27: Solve $f(x)=2$ for $f(x)=\begin{cases}x^2,&x<0\\x+2,&x\ge^0\end{cases}$.

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

Flashcard 28: Evaluate $|{-3}|$ using the piecewise definition of $|x|$.

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

Flashcard 29: What is the correct way to solve $f(x)=k$ for a piecewise function?

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

Flashcard 30: Identify the sub-function for $x < -1$ in $f(x) = \begin{cases} 3x + 2 & \text{if } x < -1 \\ x^2 + 1 & \text{if } x \geq -1 \end{cases}$.

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

Flashcard 31: What is the first step in evaluating a piecewise function?

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

Flashcard 32: Find $f(2)$ for $f(x) = \begin{cases} 5 & \text{if } x < 1 \\ x^2 & \text{if } x \geq 1 \end{cases}$.

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

Flashcard 33: Find $f(0)$ for $f(x) = \begin{cases} x^2 - 3 & \text{if } x < 0 \\ 3x + 4 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 34: What point style on a graph represents an included endpoint, $\le$ or $\ge$?

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

Flashcard 35: What is a common application of piecewise functions?

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

Flashcard 36: Find $c$ so $f$ is continuous at $x=1$: $f(x)=\begin{cases}3x+c,&x<1\\x^2,&x\ge^1\end{cases}$.

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

Flashcard 37: Evaluate $f(3)$ for $f(x)=\begin{cases}4,&x<1\\x-2,&x\ge^1\end{cases}$.

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

Flashcard 38: Find $f(3)$ for $f(x) = \begin{cases} x + 3 & \text{if } x < 3 \\ 5x - 1 & \text{if } x \geq 3 \end{cases}$.

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

Flashcard 39: Identify the sub-function for $x \geq 0$ in $f(x) = \begin{cases} x^3 & \text{if } x < 0 \\ 4x & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 40: Solve $f(x)=0$ for $f(x)=\begin{cases}x+3,&x\le-1\\x-1,&x>-1\end{cases}$.

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

Flashcard 41: What is the notation for a piecewise function?

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

Flashcard 42: Identify the interval notation for the condition $-2<x\le^5$.

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

Flashcard 43: Determine $f(-2)$ for $f(x) = \begin{cases} x^2 + 4 & \text{if } x < 0 \\ x + 5 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 44: Identify the correct endpoint inclusion for $x\in[1,3)$ in inequality form.

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

Flashcard 45: Solve $f(x)=4$ for $f(x)=\begin{cases}x+2,&x<1\\2x,&x\ge^1\end{cases}$.

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

Flashcard 46: Identify $f(0)$ for $f(x)=\begin{cases}1,&x<0\\2,&x>0\end{cases}$.

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

Flashcard 47: Identify the sub-function for $x < 2$ in $f(x) = \begin{cases} x^3 - x & \text{if } x < 2 \\ 2x^2 & \text{if } x \geq 2 \end{cases}$.

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

Flashcard 48: What is the purpose of using piecewise functions?

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

Flashcard 49: Find $a$ so the pieces match at $x=3$: $f(x)=\begin{cases}x+a,&x<3\\2x-1,&x\ge^3\end{cases}$.

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

Flashcard 50: Evaluate $|{-3}|$ using the piecewise definition of $|x|$.

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

Flashcard 51: What is the key feature of a piecewise function?

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

Flashcard 52: Identify the sub-function applied at $x = 0$ in $f(x) = \begin{cases} -x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 53: Identify the sub-function for $x < 2$ in $f(x) = \begin{cases} x^3 - x & \text{if } x < 2 \\ 2x^2 & \text{if } x \geq 2 \end{cases}$.

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

Flashcard 54: Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.

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

Flashcard 55: What is an essential property of a piecewise function's graph?

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

Flashcard 56: Find $f(0)$ for $f(x) = \begin{cases} x^2 - 3 & \text{if } x < 0 \\ 3x + 4 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 57: Solve $f(x)=2$ for $f(x)=\begin{cases}x^2,&x<0\\x+2,&x\ge^0\end{cases}$.

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

Flashcard 58: Evaluate $f(2)$ for $f(x)=\begin{cases}x+1,&x<2\\3x,&x\ge^2\end{cases}$.

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

Flashcard 59: Evaluate $f(3)$ for $f(x)=\begin{cases}4,&x<1\\x-2,&x\ge^1\end{cases}$.

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

Flashcard 60: Evaluate $f(0)$ for $f(x)=\begin{cases}5-x,&x<0\\x^2+2,&x\ge^0\end{cases}$.

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

Flashcard 61: Rewrite $f(x)=|x|+1$ as a piecewise function.

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

Flashcard 62: What point style on a graph represents an excluded endpoint, $<$ or $>$?

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

Flashcard 63: Identify the sub-function at $x = 4$ for $f(x) = \begin{cases} x^3 & \text{if } x < 4 \\ 2x + 5 & \text{if } x \geq 4 \end{cases}$.

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

Flashcard 64: What is the domain of a piecewise function?

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

Flashcard 65: What must be true for a piecewise function to be continuous?

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

Flashcard 66: What is the absolute value identity as a piecewise definition for $|x|$?

Answer: $$. Definition splits at zero where sign changes.

Flashcard 67: Evaluate $f(5)$ for $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$.

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

Flashcard 68: Find $f(0)$ for $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 3x + 2 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 69: Identify $f(0)$ for $f(x)=\begin{cases}1,&x<0\\2,&x>0\end{cases}$.

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

Flashcard 70: Find $b$ so $f$ is continuous at $x=-1$: $f(x)=\begin{cases}x^2+b,&x\le-1\\2x,&x>-1\end{cases}$.

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

Flashcard 71: What is a piecewise function?

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

Flashcard 72: What is an example of a piecewise function?

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

Flashcard 73: Evaluate $f(-1)$ for $f(x)=\begin{cases}x^2,&x\le^0\\2x+1,&x>0\end{cases}$.

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

Flashcard 74: Which branch gives $f(5)$ if $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$?

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

Flashcard 75: What is the first step in evaluating a piecewise function?

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

Flashcard 76: What is the notation for a piecewise function?

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

Flashcard 77: Determine $f(1)$ for $f(x) = \begin{cases} 7 - x & \text{if } x < 1 \\ 2x & \text{if } x \geq 1 \end{cases}$.

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

Flashcard 78: Identify the range of $f(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ 2x - 3 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 79: What is the range of a piecewise function?

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

Flashcard 80: Which branch gives $f(5)$ if $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$?

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

Flashcard 81: Find $f(2)$ for $f(x) = \begin{cases} 5 & \text{if } x < 1 \\ x^2 & \text{if } x \geq 1 \end{cases}$.

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

Flashcard 82: Evaluate $f(2)$ for $f(x)=\begin{cases}x+1,&x<2\\3x,&x\ge^2\end{cases}$.

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

Flashcard 83: How do you identify sub-functions in a piecewise function?

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

Flashcard 84: Identify whether $f$ is continuous at $x=1$: $f(x)=\begin{cases}x+4,&x<1\\x^2,&x\ge^1\end{cases}$.

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

Flashcard 85: Determine $f(-4)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ -x + 2 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 86: Determine $f(-3)$ for $f(x) = \begin{cases} 2x + 1 & \text{if } x \leq 0 \\ x^2 - 4 & \text{if } x > 0 \end{cases}$.

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

Flashcard 87: What is a characteristic of the transition between sub-functions in a piecewise function?

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

Flashcard 88: Identify the sub-function for $x \geq 0$ in $f(x) = \begin{cases} x^3 & \text{if } x < 0 \\ 4x & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 89: Determine $f(-3)$ for $f(x) = \begin{cases} 2x + 1 & \text{if } x \leq 0 \\ x^2 - 4 & \text{if } x > 0 \end{cases}$.

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

Flashcard 90: Determine $f(-2)$ for $f(x) = \begin{cases} x^2 + 4 & \text{if } x < 0 \\ x + 5 & \text{if } x \geq 0 \end{cases}$.

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

Flashcard 91: What is the meaning of $f(c)$ when the piecewise definition includes $x=c$ in one branch?

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

Flashcard 92: What is the key challenge in graphing piecewise functions?

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

Flashcard 93: Find $f(4)$ for $f(x) = \begin{cases} 2x - 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}$.

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

Flashcard 94: Evaluate $f(0)$ for $f(x)=\begin{cases}5-x,&x<0\\x^2+2,&x\ge^0\end{cases}$.

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

Flashcard 95: What does each sub-function in a piecewise function have?

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

Flashcard 96: Identify the sub-function for $x \geq 3$ in $f(x) = \begin{cases} x + 2 & \text{if } x < 3 \\ 4x & \text{if } x \geq 3 \end{cases}$.

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

Flashcard 97: Find the value at the boundary: $f(1)$ for $f(x)=\begin{cases}x+4,&x<1\\x^2,&x\ge^1\end{cases}$.

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

Flashcard 98: Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.

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

Flashcard 99: Identify which inequality symbol, $<$ or $\le$, includes the endpoint in an interval.

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

Flashcard 100: What is the absolute value identity as a piecewise definition for $|x|$?

Answer: $|x|=\begin{cases}x,&x\ge^0\\-x,&x<0\end{cases}$. Definition splits at zero where sign changes.