which table represents an exponential function of the form $y = b^x$ when $0 < b < 1$?
which table represents an exponential function of the form $y = b^x$ when $0 < b < 1$?
Answer
Explanation:
Step1: Recall the properties of (y = b^{x}) when (0 < b < 1)
When (x = 0), (y=b^{0}=1). When (x) increases, (y) decreases (since (0 < b<1), for example, if (b=\frac{1}{3}), (y = (\frac{1}{3})^{x}=3^{-x})). When (x) decreases (i.e., (x=-n,n>0)), (y = b^{-n}=\frac{1}{b^{n}}) and (y) increases.
Step2: Analyze the first table
For the first table, when (x) increases from (- 3) to (3), (y) values (\frac{1}{27},\frac{1}{9},\frac{1}{3},1,3,9,27) are increasing. If (y = b^{x}), using (x = 1,y = 3), then (b = 3>1).
Step3: Analyze the second table
For the second table, when (x=-3,y = 27); (x=-2,y = 9); (x=-1,y = 3); (x = 0,y = 1); (x = 1,y=\frac{1}{3}); (x = 2,y=\frac{1}{9}); (x = 3,y=\frac{1}{27}). Let's use the formula (y=b^{x}). When (x = 1), (y=\frac{1}{3}), so (b=\frac{1}{3}) (since (y=b^{x}) and substituting (x = 1,y=\frac{1}{3}) gives (b=\frac{1}{3}) which satisfies (0 < b<1)). Also, when (x=-1), (y=b^{-1}=\frac{1}{b}=3) (consistent with (b = \frac{1}{3})), when (x = 2), (y=b^{2}=(\frac{1}{3})^{2}=\frac{1}{9})
Answer:
The second table.