F-LE Exponential growth versus polynomial growth

(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.

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.