Piecewise Functions - ACT Math
Card 1 of 30
Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.
Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.
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$5$. Since $1 \geq 1$, use $2x^2 + 3$: $2(1)^2 + 3 = 5$.
$5$. Since $1 \geq 1$, use $2x^2 + 3$: $2(1)^2 + 3 = 5$.
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What is the correct way to solve $f(x)=g(x)$ when $f$ is piecewise?
What is the correct way to solve $f(x)=g(x)$ when $f$ is piecewise?
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Set each branch equal to $g(x)$ and restrict solutions to that branch interval. Apply each piece separately and verify domain restrictions.
Set each branch equal to $g(x)$ and restrict solutions to that branch interval. Apply each piece separately and verify domain restrictions.
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Evaluate $f(-1)$ for $f(x)=\begin{cases}x^2,&x\le^0\\2x+1,&x>0\end{cases}$.
Evaluate $f(-1)$ for $f(x)=\begin{cases}x^2,&x\le^0\\2x+1,&x>0\end{cases}$.
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$1$. Since $-1 \le 0$, use $x^2 = (-1)^2 = 1$.
$1$. Since $-1 \le 0$, use $x^2 = (-1)^2 = 1$.
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Evaluate $f(0)$ for $f(x)=\begin{cases}5-x,&x<0\\x^2+2,&x\ge^0\end{cases}$.
Evaluate $f(0)$ for $f(x)=\begin{cases}5-x,&x<0\\x^2+2,&x\ge^0\end{cases}$.
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$2$. Since $0 \ge 0$, use $x^2 + 2 = 0 + 2 = 2$.
$2$. Since $0 \ge 0$, use $x^2 + 2 = 0 + 2 = 2$.
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Evaluate $f(2)$ for $f(x)=\begin{cases}x+1,&x<2\\3x,&x\ge^2\end{cases}$.
Evaluate $f(2)$ for $f(x)=\begin{cases}x+1,&x<2\\3x,&x\ge^2\end{cases}$.
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$6$. Since $2 \ge 2$, use $3x = 3(2) = 6$.
$6$. Since $2 \ge 2$, use $3x = 3(2) = 6$.
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Evaluate $f(-2)$ for $f(x)=\begin{cases}2x+3,&x\le-1\\x^2,&x>-1\end{cases}$.
Evaluate $f(-2)$ for $f(x)=\begin{cases}2x+3,&x\le-1\\x^2,&x>-1\end{cases}$.
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$-1$. Since $-2 \le -1$, use $2x + 3 = 2(-2) + 3 = -1$.
$-1$. Since $-2 \le -1$, use $2x + 3 = 2(-2) + 3 = -1$.
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Identify the sub-function for $x \geq 3$ in $f(x) = \begin{cases} x + 2 & \text{if } x < 3 \\ 4x & \text{if } x \geq 3 \end{cases}$.
Identify the sub-function for $x \geq 3$ in $f(x) = \begin{cases} x + 2 & \text{if } x < 3 \\ 4x & \text{if } x \geq 3 \end{cases}$.
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$4x$. This is the second sub-function in the definition.
$4x$. This is the second sub-function in the definition.
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Find $f(0)$ for $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 3x + 2 & \text{if } x \geq 0 \end{cases}$.
Find $f(0)$ for $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 3x + 2 & \text{if } x \geq 0 \end{cases}$.
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$2$. Since $0 \geq 0$, use $3x + 2$: $3(0) + 2 = 2$.
$2$. Since $0 \geq 0$, use $3x + 2$: $3(0) + 2 = 2$.
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Solve $f(x)=1$ for $f(x)=\begin{cases}2-x,&x<1\\x^2,&x\ge^1\end{cases}$.
Solve $f(x)=1$ for $f(x)=\begin{cases}2-x,&x<1\\x^2,&x\ge^1\end{cases}$.
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$x=1$. Since $x \ge 1$, solve $x^2 = 1$ giving $x = 1$.
$x=1$. Since $x \ge 1$, solve $x^2 = 1$ giving $x = 1$.
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Which branch gives $f(5)$ if $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$?
Which branch gives $f(5)$ if $f(x)=\begin{cases}x-1,&x<0\\x+1,&0\le x<4\\2x,&x\ge^4\end{cases}$?
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Use $2x$ because $x\ge^4$. Check which condition contains $x = 5$.
Use $2x$ because $x\ge^4$. Check which condition contains $x = 5$.
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Evaluate $|{-3}|$ using the piecewise definition of $|x|$.
Evaluate $|{-3}|$ using the piecewise definition of $|x|$.
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$3$. Since $-3 < 0$, use $-x = -(-3) = 3$.
$3$. Since $-3 < 0$, use $-x = -(-3) = 3$.
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Find $k$ so $f$ is continuous at $x=0$: $f(x)=\begin{cases}k-x,&x<0\\x+2,&x\ge^0\end{cases}$.
Find $k$ so $f$ is continuous at $x=0$: $f(x)=\begin{cases}k-x,&x<0\\x+2,&x\ge^0\end{cases}$.
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$k=2$. Set left limit $k = 2$ to match right value.
$k=2$. Set left limit $k = 2$ to match right value.
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What is the first step in evaluating a piecewise function?
What is the first step in evaluating a piecewise function?
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Determine which sub-function applies to the given input value. Check which condition the input satisfies first.
Determine which sub-function applies to the given input value. Check which condition the input satisfies first.
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Find $b$ so $f$ is continuous at $x=-1$: $f(x)=\begin{cases}x^2+b,&x\le-1\\2x,&x>-1\end{cases}$.
Find $b$ so $f$ is continuous at $x=-1$: $f(x)=\begin{cases}x^2+b,&x\le-1\\2x,&x>-1\end{cases}$.
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$b=-3$. Set left value $1 + b = -2$ to match right limit.
$b=-3$. Set left value $1 + b = -2$ to match right limit.
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What is the absolute value identity as a piecewise definition for $|x|$?
What is the absolute value identity as a piecewise definition for $|x|$?
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$|x|=\begin{cases}x,&x\ge^0\\-x,&x<0\end{cases}$. Definition splits at zero where sign changes.
$|x|=\begin{cases}x,&x\ge^0\\-x,&x<0\end{cases}$. Definition splits at zero where sign changes.
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What is the piecewise definition of $|x-a|$?
What is the piecewise definition of $|x-a|$?
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$|x-a|=\begin{cases}x-a,&x\ge a\\a-x,&x<a\end{cases}$. Expression inside absolute value changes sign at $x=a$.
$|x-a|=\begin{cases}x-a,&x\ge a\\a-x,&x<a\end{cases}$. Expression inside absolute value changes sign at $x=a$.
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Identify $f(0)$ for $f(x)=\begin{cases}1,&x<0\\2,&x>0\end{cases}$.
Identify $f(0)$ for $f(x)=\begin{cases}1,&x<0\\2,&x>0\end{cases}$.
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$f(0)$ is undefined. No piece includes $x = 0$ in its domain.
$f(0)$ is undefined. No piece includes $x = 0$ in its domain.
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Solve $f(x)=0$ for $f(x)=\begin{cases}x+3,&x\le-1\\x-1,&x>-1\end{cases}$.
Solve $f(x)=0$ for $f(x)=\begin{cases}x+3,&x\le-1\\x-1,&x>-1\end{cases}$.
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$x=-3$ and $x=1$. Solve in both pieces: $x + 3 = 0$ and $x - 1 = 0$.
$x=-3$ and $x=1$. Solve in both pieces: $x + 3 = 0$ and $x - 1 = 0$.
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What is a piecewise function?
What is a piecewise function?
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A function defined by multiple sub-functions, each applying to a certain interval. Different rules apply to different input ranges.
A function defined by multiple sub-functions, each applying to a certain interval. Different rules apply to different input ranges.
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Determine $f(6)$ for $f(x) = \begin{cases} 2x - 5 & \text{if } x < 4 \\ x^2 + 1 & \text{if } x \geq 4 \end{cases}$.
Determine $f(6)$ for $f(x) = \begin{cases} 2x - 5 & \text{if } x < 4 \\ x^2 + 1 & \text{if } x \geq 4 \end{cases}$.
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$37$. Since $6 \geq 4$, use $x^2 + 1$: $6^2 + 1 = 37$.
$37$. Since $6 \geq 4$, use $x^2 + 1$: $6^2 + 1 = 37$.
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Find $f(4)$ for $f(x) = \begin{cases} 2x - 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}$.
Find $f(4)$ for $f(x) = \begin{cases} 2x - 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}$.
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$14$. Since $4 \geq 3$, use $x^2 - 2$: $4^2 - 2 = 14$.
$14$. Since $4 \geq 3$, use $x^2 - 2$: $4^2 - 2 = 14$.
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What is an essential property of a piecewise function's graph?
What is an essential property of a piecewise function's graph?
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It may have different slopes or curvatures in different intervals. Each piece can have distinct mathematical behavior.
It may have different slopes or curvatures in different intervals. Each piece can have distinct mathematical behavior.
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Identify the sub-function applied at $x = 0$ in $f(x) = \begin{cases} -x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$.
Identify the sub-function applied at $x = 0$ in $f(x) = \begin{cases} -x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$.
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$x^2$. Since $0 \geq 0$, the second sub-function applies.
$x^2$. Since $0 \geq 0$, the second sub-function applies.
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Determine $f(-4)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ -x + 2 & \text{if } x \geq 0 \end{cases}$.
Determine $f(-4)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ -x + 2 & \text{if } x \geq 0 \end{cases}$.
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$16$. Since $-4 < 0$, use $x^2$: $(-4)^2 = 16$.
$16$. Since $-4 < 0$, use $x^2$: $(-4)^2 = 16$.
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What is the key challenge in graphing piecewise functions?
What is the key challenge in graphing piecewise functions?
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Ensuring correct transitions and connections between sub-functions. Must handle discontinuities and different function behaviors.
Ensuring correct transitions and connections between sub-functions. Must handle discontinuities and different function behaviors.
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Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.
Find $f(1)$ for $f(x) = \begin{cases} x - 2 & \text{if } x < 1 \\ 2x^2 + 3 & \text{if } x \geq 1 \end{cases}$.
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$5$. Since $1 \geq 1$, use $2x^2 + 3$: $2(1)^2 + 3 = 5$.
$5$. Since $1 \geq 1$, use $2x^2 + 3$: $2(1)^2 + 3 = 5$.
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Identify the sub-function for $x < -1$ in $f(x) = \begin{cases} 3x + 2 & \text{if } x < -1 \\ x^2 + 1 & \text{if } x \geq -1 \end{cases}$.
Identify the sub-function for $x < -1$ in $f(x) = \begin{cases} 3x + 2 & \text{if } x < -1 \\ x^2 + 1 & \text{if } x \geq -1 \end{cases}$.
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$3x + 2$. This is the first sub-function for the given condition.
$3x + 2$. This is the first sub-function for the given condition.
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What is a characteristic of the transition between sub-functions in a piecewise function?
What is a characteristic of the transition between sub-functions in a piecewise function?
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The transition may be continuous or discontinuous. Depends on whether boundary values align between pieces.
The transition may be continuous or discontinuous. Depends on whether boundary values align between pieces.
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Determine $f(0)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases}$.
Determine $f(0)$ for $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases}$.
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$1$. Since $0 \geq 0$, use $x + 1$: $0 + 1 = 1$.
$1$. Since $0 \geq 0$, use $x + 1$: $0 + 1 = 1$.
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What must be true for a piecewise function to be continuous?
What must be true for a piecewise function to be continuous?
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Sub-functions must connect at their boundaries without jumps. Function values must match at boundary points.
Sub-functions must connect at their boundaries without jumps. Function values must match at boundary points.
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