Piecewise Functions
Help Questions
ACT Math › Piecewise Functions
Which interval contains $x = -1$ for the piecewise function $f(x) = \begin{cases} 3x + 7 & \text{if } x \leq -1 \\ x^2 - 2 & \text{if } x > -1 \end{cases}$?
$x \leq -1$
$x < -1$
$x = 0$
$x > -1$
Explanation
To determine which interval contains $x = -1$, we check each condition: Is $-1 \leq -1$? Yes. Is $-1 > -1$? No. Since $-1$ satisfies the condition $x \leq -1$, it belongs to the first interval. The boundary point $x = -1$ is included in the first piece due to the $\leq$ symbol.
What is $f(0)$ for the piecewise function $f(x) = \begin{cases} 2x + 4 & \text{if } x < 1 \\ x^2 - 6 & \text{if } x \geq 1 \end{cases}$?
0
2
4
6
Explanation
For x = 0, we check the intervals: Is 0 < 1? Yes. So we use the first piece: $f(x) = 2x + 4$. Substituting x = 0: $f(0) = 2(0) + 4 = 0 + 4 = 4$. Choice B would result from using the second piece incorrectly.
A savings plan applies a rule $f(x)$ to the number of weeks $x$ you have saved. For the piecewise function $$f(x)=\begin{cases}6-x & \text{if } x<4\\ 2x+1 & \text{if } 4\le x<9\\ x^2-10 & \text{if } x\ge 9\end{cases}$$ what is $f(9)$?
8
19
71
81
Explanation
For x = 9, determine which piece to use: Is 9 < 4? No. Is 4 ≤ 9 < 9? No, since 9 is not less than 9. Is 9 ≥ 9? Yes. Use the third piece: f(x) = x² - 10. Thus f(9) = 9² - 10 = 81 - 10 = 71.
For the piecewise function $$f(x) = \begin{cases} x + 3 & \text{if } x < -2 \\ -2x & \text{if } -2 \leq x < 3 \\ x^2 - 5 & \text{if } x \geq 3 \end{cases}$$, what is f(3)?
3
4
5
9
Explanation
For $x = 3$, check intervals: $3 < -2$? No. $-2 \leq 3 < 3$? No. $3 \geq 3$? Yes. So use the third piece $f(x) = x^2 - 5$. Substitute $x = 3$: $f(3) = 3^2 - 5 = 9 - 5 = 4$. Note that $x = 3$ falls in the third piece due to the $\geq$ condition.
What is f(1) for the piecewise function: $$f(x) = \begin{cases} x^2 - 1 & \text{if } x < 1 \\ 3 & \text{if } 1 \leq x < 4 \\ 4x & \text{if } x \geq 4 \end{cases}$$?
0
1
3
4
Explanation
For x = 1, check intervals: 1 < 1? No. 1 ≤ 1 < 4? Yes. So use the second piece $f(x) = 3$. At the boundary x = 1, we use the second piece due to the ≤ condition. Therefore $f(1) = 3$.
Which interval contains x = 3 for the function $$f(x) = \begin{cases} 3x + 1 & \text{if } x < 1 \\ 2x - 2 & \text{if } 1 \leq x < 4 \\ x^2 & \text{if } x \geq 4 \end{cases}$$?
$x > 4$
$x < 1$
$x \geq 4$
$1 \leq x < 4$
Explanation
For $x = 3$, check each interval: $3 < 1$? No. $1 \leq 3 < 4$? Yes, since $1 \leq 3$ and $3 < 4$. $3 \geq 4$? No. Therefore, $x = 3$ falls in the interval $1 \leq x < 4$.
Based on the piecewise function $$f(x) = \begin{cases} 2x - 3 & \text{if } x < -1 \\ x^2 + 2 & \text{if } -1 \leq x < 2 \\ 5x - 1 & \text{if } x \geq 2 \end{cases}$$, what is f(2)?
3
4
9
10
Explanation
For x = 2, check intervals: $2 < -1$? No. $-1 \leq 2 < 2$? No. $2 \geq 2$? Yes. So use the third piece $f(x) = 5x - 1$. Substitute x = 2: $f(2) = 5(2) - 1 = 10 - 1 = 9$. Note that x = 2 falls in the third piece due to the $\geq$ condition.
For the piecewise function $$f(x) = \begin{cases} 3x^2 & \text{if } x < -1 \\ x - 5 & \text{if } -1 \leq x < 2 \\ 2x + 3 & \text{if } x \geq 2 \end{cases}$$, what is f(2)?
6
7
8
9
Explanation
For x = 2, check intervals: 2 < -1? No. -1 ≤ 2 < 2? No. 2 ≥ 2? Yes. So use the third piece $f(x) = 2x + 3$. Substitute x = 2: $f(2) = 2(2) + 3 = 4 + 3 = 7$. Note that x = 2 falls in the third piece due to the ≥ condition.
What is f(2) for the piecewise function: $$f(x) = \begin{cases} -x + 3 & \text{if } x < 1 \\ 4x & \text{if } 1 \leq x < 3 \\ x^2 - 1 & \text{if } x \geq 3 \end{cases}$$?
6
7
8
9
Explanation
For x = 2, check intervals: 2 < 1? No. 1 ≤ 2 < 3? Yes. So use the second piece $f(x) = 4x$. Substitute x = 2: $f(2) = 4(2) = 8$. The value x = 2 falls clearly within the middle interval.
A company assigns a performance rating $f(x)$ based on an employee’s score $x$. The rating function is
$$f(x)=\begin{cases}
7-x & \text{if } x<0 \
3x+1 & \text{if } 0\le x<4 \
15 & \text{if } x\ge 4
\end{cases}$$
Based on the piecewise function, what is the value when $x=0$?
0
1
7
15
Explanation
For x = 0, check intervals: Is 0 < 0? No. Is 0 ≤ 0 < 4? Yes, since 0 = 0 satisfies this condition. Use the second piece: f(x) = 3x + 1. Substituting: f(0) = 3(0) + 1 = 0 + 1 = 1. The boundary x = 0 falls in the middle piece due to the ≤ sign.