This principle can be proven by considering the function ${\displaystyle h(x)=f(x)-g(x)}$ . If we were to take the derivative we would notice that for ${\displaystyle x>0}$ ,

${\displaystyle h'=f'-g'>0.}$ 

Also notice that ${\displaystyle h(0)=0}$ . Combining these observations, we can use the mean value theorem on the interval ${\displaystyle [0,x]}$  and get

${\displaystyle 0<h'(x_{0})={\frac {h(x)-h(0)}{x-0}}={\frac {f(x)-g(x)}{x}}.}$ 

By assumption, ${\displaystyle x>0}$ , so multiplying both sides by ${\displaystyle x}$  gives ${\displaystyle f(x)-g(x)>0}$ . This implies ${\displaystyle f(x)>g(x)}$ .

The statement of the racetrack principle can slightly generalized as follows;

if ${\displaystyle f'(x)>g'(x)}$  for all ${\displaystyle x>a}$ , and if ${\displaystyle f(a)=g(a)}$ , then ${\displaystyle f(x)>g(x)}$  for all ${\displaystyle x>a}$ .

as above, substituting ≥ for > produces the theorem

if ${\displaystyle f'(x)\geq g'(x)}$  for all ${\displaystyle x>a}$ , and if ${\displaystyle f(a)=g(a)}$ , then ${\displaystyle f(x)\geq g(x)}$  for all ${\displaystyle x>a}$ .

Proof edit

This generalization can be proved from the racetrack principle as follows:

Consider functions ${\displaystyle f_{2}(x)=f(x+a)}$  and ${\displaystyle g_{2}(x)=g(x+a)}$ . Given that ${\displaystyle f'(x)>g'(x)}$  for all ${\displaystyle x>a}$ , and ${\displaystyle f(a)=g(a)}$ ,

${\displaystyle f_{2}'(x)>g_{2}'(x)}$  for all ${\displaystyle x>0}$ , and ${\displaystyle f_{2}(0)=g_{2}(0)}$ , which by the proof of the racetrack principle above means ${\displaystyle f_{2}(x)>g_{2}(x)}$  for all ${\displaystyle x>0}$  so ${\displaystyle f(x)>g(x)}$  for all ${\displaystyle x>a}$ .

The racetrack principle can be used to prove a lemma necessary to show that the exponential function grows faster than any power function. The lemma required is that

${\displaystyle e^{x}>x}$ 

for all real ${\displaystyle x}$ . This is obvious for ${\displaystyle x<0}$  but the racetrack principle is required for ${\displaystyle x>0}$ . To see how it is used we consider the functions

${\displaystyle f(x)=e^{x}}$ 

and

${\displaystyle g(x)=x+1.}$ 

Notice that ${\displaystyle f(0)=g(0)}$  and that

${\displaystyle e^{x}>1}$ 

because the exponential function is always increasing (monotonic) so ${\displaystyle f'(x)>g'(x)}$ . Thus by the racetrack principle ${\displaystyle f(x)>g(x)}$ . Thus,

${\displaystyle e^{x}>x+1>x}$ 

for all ${\displaystyle x>0}$ .

• Deborah Hughes-Hallet, et al., Calculus.