Calculus- Early Transcendentals, 2021a

54 Functions
After the positive integers, the next easiest number to understand is 0: 2 0 = 1. You have presumably
learned this fact in the past; why is it true? It is true precisely because we want 2 a 2 b = 2 a+b to be true
about the function 2 x . We need it to be true that 2 0 2 x = 2 0+x = 2 x , and this only works if 2 0 = 1. The same
argument implies that a 0 = 1foranya.
The next easiest set of numbers to understand is the negative integers: for example, 2 −3 = 1/2 3 .We
know that whatever 2 −3 meansitmustbethat2 −3 2 3 = 2 −3+3 = 2 0 = 1, which means that 2 −3 must be
1/2 3 . In fact, by the same argument, once we know what 2 x means for some value of x, 2 −x must be 1/2 x
and more generally a −x = 1/a x .
Next, consider an exponent 1/q, whereq is a positive integer. We want it to be true that (2 x ) y = 2 xy ,
so (2 1/q ) q = 2. This means that 2 1/q is a q-th root of 2, 2 1/q = q√ 2 . This is all we need to understand that
2 p/q =(2 1/q ) p =( q√ 2 ) p and a p/q =(a 1/q ) p =( q√ a ) p .
What’s left is the hard part: what does 2 x mean when x cannot be written as a fraction, like x = √ 2or
x = π? What we know so far is how to assign meaning to 2 x whenever x = p/q. Ifweweretographa x
(for some a > 1) at points x = p/q then we’d see something like this:
This is a poor picture, but it illustrates a series of individual points above the rational numbers on the
x-axis. There are really a lot of “holes” in the curve, above x = π, for example. But (this is the hard part)
it is possible to prove that the holes can be “filled in”, and that the resulting function, called a x , really does
have the properties we want, namely that a x a y = a x+y and (a x ) y = a xy . Such a graph would then look like
this: