Proving f(x) > e^x for a Monotonic Function with f'(x) > f(x) and f(0) = 1

[talolard]

Homework Statement

Let f be a function from R to R such that f'(x)>f(x), f(0)=1 for every x. prove that for every x>0 f(x)>e^x

The Attempt at a Solution

From the given information we know that f'(0) \geq 0 and so f'(x) \geq 0

define h(x) = f(x)-e^x
so h(0) =0

h'(x)=f'(x) -e^x = lim_{h ->0 } \frac {f(x+h)-f(x)}{h} -e^x= lim_{h ->0 } \frac {f(x+h)-f(x) -he^x}{h}= lim_{h ->0 } \frac {f(x+h)-f(x) -0}{h} = f'(x) \geq 0

Then h(x) is montonous rising and so is never smaller then zero. thus f(x) \geq e^xIs this correct?

Thanks
Tal

 

[snipez90]

Well you seemed to have defined f(0) = 0, so that h(0) = -1...

 

[talolard]

oops. I made a typo, that was supposed to be f(0)=1. Fixed in the original post.
Thanks

 

[snipez90]

OK cool. Even though you have the right idea, the second to last equality in your main derivation cannot be justified, since you are taking the limit of some term in the numerator of a quotient which is a serious violation of the limit laws (basically you need to know all limits in the the numerator and denominator exist before you can perform such operations).

My advice would be to consider g(x) = f(x)/e^x instead and show g is increasing. At no point should you need to resort to the definition of the derivative.

 

[talolard]

That seems much simpler:
Just to make sure i got it right:
h(x) = \frac {f(x)}{e^x}, h(0)=1
h'(x)= \frac {f'(x)e^x -f(x)e^x}{e^(2x)} \geq 0 because f'(x)>f(x)
Thanks
Tal

 

[snipez90]

That's fine. Notice the last inequality is in fact strict, which only helps you I guess.

 

[talolard]

Thanks alot. Much apreciated