Racetrack principle

Incalculus, theracetrack principledescribes the movement and growth of two functions in terms of theirderivatives.

This principle is derived from the fact that if a horse named Frank Fleetfeet always runs faster than a horse named Greg Gooseleg, then if Frank and Greg start a race from the same place and the same time, then Frank will win. More briefly, the horse that starts fast and stays fast wins.

In symbols:

if$f'(x)>g'(x)$for all$x>0$,and if$f(0)=g(0)$,then$f(x)>g(x)$for all$x>0$.

or, substituting ≥ for > produces the theorem

if$f'(x)\geq g'(x)$for all$x>0$,and if$f(0)=g(0)$,then$f(x)\geq g(x)$for all$x\geq 0$.

which can be proved in a similar way

Proof

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

$h'=f'-g'>0.$

Also notice that$h(0)=0$.Combining these observations, we can use themean value theoremon the interval$[0,x]$and get

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

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

Generalizations

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

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

as above, substituting ≥ for > produces the theorem

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

Proof

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

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

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

Application

The racetrack principle can be used to prove alemmanecessary to show that theexponential functiongrows faster than any power function. The lemma required is that

$e^{x}>x$

for all real$x$.This is obvious for$x<0$but the racetrack principle can be used for$x>0$.To see how it is used we consider the functions

$f(x)=e^{x}$

and

$g(x)=x+1.$

Notice that$f(0)=g(0)$and that

$e^{x}>1$

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

$e^{x}>x+1>x$

for all$x>0$.

References

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