Question

consider the following function. f(x) = (7)·2^{x - 3} enter the missing values in the table below in the order they would appear left - to - right separated by a comma. \

$$\begin{tabular}{|c|c|c|c|c|c|} \\hline x & -1 & 0 & 1 & 2 & 3 \\\\ \\hline f(x) & & & & & \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Calculate for x=-1

Substitute \( x = -1 \) into \( f(x) = 7 \cdot 2^{x - 3} \).
\( f(-1)=7\cdot2^{-1 - 3}=7\cdot2^{-4}=7\cdot\frac{1}{16}=\frac{7}{16}=0.4375 \)

Step2: Calculate for x=0

Substitute \( x = 0 \) into \( f(x) \).
\( f(0)=7\cdot2^{0 - 3}=7\cdot2^{-3}=7\cdot\frac{1}{8}=\frac{7}{8}=0.875 \)

Step3: Calculate for x=1

Substitute \( x = 1 \) into \( f(x) \).
\( f(1)=7\cdot2^{1 - 3}=7\cdot2^{-2}=7\cdot\frac{1}{4}=\frac{7}{4}=1.75 \)

Step4: Calculate for x=2

Substitute \( x = 2 \) into \( f(x) \).
\( f(2)=7\cdot2^{2 - 3}=7\cdot2^{-1}=7\cdot\frac{1}{2}=\frac{7}{2}=3.5 \)

Step5: Calculate for x=3

Substitute \( x = 3 \) into \( f(x) \).
\( f(3)=7\cdot2^{3 - 3}=7\cdot2^{0}=7\cdot1 = 7 \)

Answer:

0.4375, 0.875, 1.75, 3.5, 7