Question

a. $f(x) = \

$$\begin{cases} -5, & x \\leq -2 \\\\ x^2 + 5, & -2 < x < 1 \\\\ 2^{(x+2)} - 3, & x \\geq 1 \\end{cases}$$

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b. $f(x) = \

$$\begin{cases} 5, & x \\leq -2 \\\\ x^2 - 5, & -2 < x < 1 \\\\ 2^{(x-2)} - 3, & x \\geq 1 \\end{cases}$$

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c. $f(x) = \

$$\begin{cases} 5, & x \\leq -2 \\\\ x^2 + 5, & -2 < x < 1 \\\\ 2^{(x+2)} - 2, & x \\geq 1 \\end{cases}$$

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Explanation:

Step1: Check left piece (x ≤ -2)

The graph has a closed point at (-2, 5), so for $x \leq -2$, $f(x)=5$.

Step2: Check middle piece (-2 < x <1)

Test $x=0$: the graph has an open point at (0, -5). Substitute $x=0$ into $x^2 -5$: $0^2 -5 = -5$, which matches.

Step3: Check right piece (x ≥1)

Test $x=1$: the graph has a closed point at (1, -1). Substitute $x=1$ into $2^{(x-2)}-3$: $2^{(1-2)}-3 = \frac{1}{2}-3 = -1$, which matches. Test $x=4$: $2^{(4-2)}-3=4-3=1$, which matches the graph's value at x=4.

Answer:

B. $f(x) =

$$\begin{cases} 5, & x \leq -2 \\ x^2 - 5, & -2 < x < 1 \\ 2^{(x-2)} - 3, & x \geq 1 \end{cases}$$

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