Question

These tables of values represent continuous functions. In which table do the values represent an exponential function?

Answer 1

Answer:

Table B represents an exponential function.

Step-by-step explanation:

An exponential function is a function which has common ratio. Using this fact we will evaluate the functions given in the form of a table.

Table A.

f(1) = 3

f(2) = 6

f(3) = 9

Now $\frac{f(2)}{f(1)}=\frac{6}{3}=\frac{2}{1}$

$\frac{f(3)}{f(2)}=\frac{9}{6}=\frac{3}{2}$

Ratios are not equal so it's not an exponential function.

Table B.

f(1) = 2

f(2) = 6

f(3) = 18

$\frac{f(2)}{f(1)}=\frac{6}{2}=\frac{3}{1}$

$\frac{f(3)}{f(2)}=\frac{18}{6}=\frac{3}{1}$

Here ratios are same therefore it's an exponential function.

Table C.

f(1) = 10

f(2) = 22

f(3) = 34

$\frac{f(2)}{f(1)}=\frac{22}{10}=\frac{11}{5}$

$\frac{f(3)}{f(2)}=\frac{34}{22}=\frac{17}{11}$

Ratios are not equal therefore it's not an exponential function.

Table D.

f(1) = 7

f(2) = 8

f(3) = 9

$\frac{f(2)}{f(1)}=\frac{8}{7}$

$\frac{f(3)}{f(2)}=\frac{9}{8}$

Ratios are not equal so it's not an exponential function.

Therefore Table B is the correct option.

Answer 2

Answer:

The correct option is B.

Step-by-step explanation:

A function is called an exponential function if it has common ratio.

A function is called an linear function if it has common difference.

In option A.

$\frac{f(2)}{f(1)}=\frac{6}{3}=2$

$\frac{f(3)}{f(2)}=\frac{9}{6}=\frac{3}{2}$

$2\neq \frac{3}{2}$

Since the given table has different ratio, therefore it is not an exponential function. Option A is incorrect.

In option B.

$\frac{f(2)}{f(1)}=\frac{6}{2}=3$

$\frac{f(3)}{f(2)}=\frac{18}{6}=3$

$3=3$

Since the given table has common ratio, therefore it is an exponential function. Option B is correct.

In option C.

$\frac{f(2)}{f(1)}=\frac{22}{10}=\frac{11}{5}$

$\frac{f(3)}{f(2)}=\frac{34}{22}=\frac{17}{11}$

$\frac{11}{5}\neq \frac{17}{11}$

Since the given table has different ratio, therefore it is not an exponential function. Option C is incorrect.

In option D.

$\frac{f(2)}{f(1)}=\frac{8}{7}$

$\frac{f(3)}{f(2)}=\frac{9}{8}$

$\frac{8}{7}\neq \frac{9}{8}$

Since the given table has different ratio, therefore it is not an exponential function. Option D is incorrect.