F-LE Exponential growth versus polynomial growth

Alignments to Content Standards: F-LE.A.3


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

  1. 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?
  2. 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.


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 $2^x$:


(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 $$ 2^{12} > 2(12)^3 + 1. $$


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.