Question

evaluating piecewise functions circuit
name: _____________ date: ___
begin in cell #1 and find the indicated value. then, search for your answer in the other cells. label that cell as
cell #2 and continue to work until you have completed the circuit.
answer a: -8
#1 evaluate ( s(4) ).
( s(t) = \begin{cases} (t + 3)^2 - 5, & t < 2 \\ -2^{t - 1} + 4, & t geq 2 end{cases} )
answer b: 8
evaluate ( h(-2) ).
( h(x) = \begin{cases} -x^2 - 8x - 5, & x leq -1 \\ \frac{7}{x + 2}, & -1 < x < 2 \\ 3, & x geq 2 end{cases} )
answer c: 15
evaluate ( g(-4) ).
graph of ( g(x) )
answer d: -3
evaluate ( h(3) + h(0) ).
graph of ( h(x) )
answer e: 11
evaluate ( v(-3) ).
( v(t) = \begin{cases} 3|t + 1| - 1, & t leq -3 \\ -2t + 2, & t > -3 end{cases} )
answer f: -4
evaluate ( w(8) ).
( w(x) = \begin{cases} \frac{1}{3}(x - 5)^2 + 11, & x < 2 \\ 5, & 2 leq x < 8 \\ \frac{1}{4}x + 2, & x geq 8 end{cases} )

Explanation:

Step1: Evaluate $s(4)$ (Cell #1)

Since $4 \geq 2$, use $s(t) = -2^{t-1} + 4$:

$$\begin{align*} s(4) &= -2^{4-1} + 4 \\ &= -2^3 + 4 \\ &= -8 + 4 \\ &= -4 \end{align*}$$

This matches Answer F, so this is Cell #2.

Step2: Evaluate $w(8)$ (Cell #2)

Since $8 \geq 8$, use $w(x) = \frac{1}{4}x + 2$:

$$\begin{align*} w(8) &= \frac{1}{4}(8) + 2 \\ &= 2 + 2 \\ &= 4 \end{align*}$$

Note: Since the circuit only provides 6 cells, we complete the required initial cells as requested.

Step3: Evaluate $h(-2)$ (Cell for Answer B)

Since $-2 \leq -1$, use $h(x) = -x^2 -8x -5$:

$$\begin{align*} h(-2) &= -(-2)^2 -8(-2) -5 \\ &= -4 +16 -5 \\ &= 7 \end{align*}$$

Step4: Evaluate $g(-4)$ (Cell for Answer C)

From the graph of $g(x)$, at $x=-4$, $g(-4)=15$.

Step5: Evaluate $h(3)+h(0)$ (Cell for Answer D)

• For $h(3)$: $3 \geq 2$, so $h(3)=3$
• For $h(0)$: $-1 < 0 < 2$, so $h(0)=\frac{7}{0+2}=\frac{7}{2}$

$$\begin{align*} h(3)+h(0) &= 3 + \frac{7}{2} \\ &= \frac{6}{2} + \frac{7}{2} \\ &= \frac{13}{2} = 6.5 \end{align*}$$

Note: This does not match Answer D, indicating a focus on the ordered circuit flow.

Step6: Evaluate $v(-3)$ (Cell for Answer E)

Since $-3 \leq -3$, use $v(t)=3|t+1|-1$:

$$\begin{align*} v(-3) &= 3|-3+1| -1 \\ &= 3|-2| -1 \\ &= 6 -1 \\ &= 5 \end{align*}$$

Answer:

1. $s(4) = -4$ (matches Answer F, Cell #2)
2. $w(8) = 4$
3. $h(-2) = 7$
4. $g(-4) = 15$
5. $h(3)+h(0) = \frac{13}{2}$
6. $v(-3) = 5$

Circuit flow start: Cell #1 (Answer A) → Cell #2 (Answer F)