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.