```Question 172663

If {{{x=0}}}, then {{{f(0)=3^0=1}}} and {{{g(0)=0}}}.

If {{{x=1}}}, then {{{f(1)=3^1=3}}} and {{{g(1)=3}}}.

If {{{x=2}}}, then {{{f(2)=3^2=9}}} and {{{g(2)=2}}}.

If {{{x=3}}}, then {{{f(3)=3^3=27}}} and {{{g(3)=3}}}.

So we have this table of values

<table border="1"><th>x</th><th>f(x)</th><th>g(x)</th><tr><td>0</td><td>1</td><td>0</td></tr><tr><td>1</td><td>3</td><td>1</td></tr><tr><td>2</td><td>9</td><td>2</td></tr><tr><td>3</td><td>27</td><td>3</td></tr></table>

From the table, we can see that {{{g(x)}}} increments by 1 as x increments by 1. On the other hand, we can see that {{{f(x)}}} goes from 1 to 3 (a difference of 2), 3 to 9 (a difference of 6), 9 to 27 (a difference of 18), etc. So the differences between each term is: 2, 6, 18, etc....

This means that from x=0 to x=1, the average rate of change for g(x) is 2. From x=1 to x=2, the average rate of change for g(x) is 6. From x=2 to x=3, the average rate of change for f(x) is 18.

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So dividing the first average rate of change 2 by 1, we get {{{2/1=2}}}. So from x=0 to x=1, {{{f(x)}}} is growing twice as fast as {{{g(x)}}}.

Dividing the second average rate of change 6 by 1, we get {{{6/1=6}}}. So from x=1 to x=2, {{{f(x)}}} is growing six times as fast as {{{g(x)}}}.

Dividing the third average rate of change 18 by 1, we get {{{18/1=18}}}. So from x=2 to x=3, {{{f(x)}}} is growing eighteen times as fast as {{{g(x)}}}.

As you can see, the exponential function is <font size=4><b>not</b></font> growing at a constant rate. So {{{f(x)}}} cannot be growing 3 times faster than {{{g(x)}}}

Note: the function {{{f(x)=3x}}} does however grow three times faster than {{{g(x)=x}}}, but that is for another problem.

So that means that the statement is false. ```