# F-LE Exponential growth versus polynomial growth

## Task

The table below shows the values of $2^x$ and $2x^3 + 1$ for some whole number values of $x$:

$x$ | $2^x$ | $2x^3+1$ |
---|---|---|

1 | 2 | 3 |

2 | 4 | 17 |

3 | 8 | 55 |

4 | 16 | 129 |

5 | 32 | 251 |

- The numbers in the third column (values of $2x^3 + 1$) are all larger than the numbers in the second column (values of $2^x$). Does this remain true if the table is extended to include whole number values up to ten?
- Explain how you know that the values of $2^x$ will eventually exceed those of the polynomial $2x^3 + 1$. What is the smallest whole number value of $x$ for which this happens?

## IM Commentary

This problem shows that an exponential function takes larger values than a cubic polynomial function provided the input is sufficiently large.

## Solutions

Solution: Table

(a) The table can be extended for whole number values of $x$ up to $x = 10$ and the values of $2x^3 + 1$ remain larger than those for $2^x$:

$x$ | $2^x$ | $2x^3+1$ |
---|---|---|

6 | 64 | 433 |

7 | 128 | 687 |

8 | 256 | 1025 |

9 | 512 | 1459 |

10 | 1024 | 2001 |

(b) If the table is continued, for all values of $x$ up to and including 11 the polynomial $2x^3 + 1$ takes a larger value than the exponential $2^x$. But $$ 2^{12} > 2(12)^3 + 1. $$

$x$ | $2^x$ | $2x^3+1$ |
---|---|---|

11 | 2048 | 2663 |

12 | 4096 | 3457 |

We know that the exponential $2^x$ will eventually exceed in value the polynomial $2x^3 + 1$ because its base, 2, is larger than one and an exponential functions grow faster, as the size of $x$ increases, than any particular polynomial function. This is explained in greater detail in the second solution below by examining quotients of $2^x$ and $2x^3+1$ when evaluated at successive whole numbers.

Solution: 2. Abstract argument

The argument presented here does not find the smallest whole number (12) where
the value of $2^x$ first exceeds the value of $2x^3 + 1$ but rather explains
why there must be such a whole number. The argument would apply not only
to $2x^3+1$ but also to any other polynomial.

Each time the variable $x$ is increased by one unit, the exponential function
$2^x$ doubles:
$$
\frac{2^{x+1}}{2^x} = 2.
$$
For the polynomial function $2x^3 + 1$, an increase in $x$ by one unit
increases the value of the function by a factor of
$$
\frac{2(x+1)^3+1}{2x^3+1} = \frac{2x^3+6x^2+6x+7}{2x^3+1}.
$$
Unlike the exponential function, these growth factors for the polynomial
function depend on the value of $x$. Notice that as $x$ increases, the
expression
$$
\frac{2x^3+6x^2+6x+7}{2x^3+1}
$$
gets closer and closer to one (because for large positive values of $x$,
the terms $6x^2, 6x, 7,$ and $1$ influence the value of the quotient by
a small quantity). Thus, as $x$ is continually incremented by one unit,
the value of $2^x$ always doubles while value of $2x^3+1$ only increases
by a factor closer and closer to one, thereby allowing the exponential values to eventually
surpass the polynomial values.

## F-LE Exponential growth versus polynomial growth

The table below shows the values of $2^x$ and $2x^3 + 1$ for some whole number values of $x$:

$x$ | $2^x$ | $2x^3+1$ |
---|---|---|

1 | 2 | 3 |

2 | 4 | 17 |

3 | 8 | 55 |

4 | 16 | 129 |

5 | 32 | 251 |

- The numbers in the third column (values of $2x^3 + 1$) are all larger than the numbers in the second column (values of $2^x$). Does this remain true if the table is extended to include whole number values up to ten?
- Explain how you know that the values of $2^x$ will eventually exceed those of the polynomial $2x^3 + 1$. What is the smallest whole number value of $x$ for which this happens?

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