Derivatives and exponential function

[icystrike]

Homework Statement

I don't understand why they took M(e)=1 , and how the proceed on with the proof.
Thanks in advance(=

Homework Equations

The Attempt at a Solution

 

[3029298]

Try to write the derivative of the exponential with the limit definition:

f'(x)=lim h->0 (f(x+h)-f(x))/h

and try to factor out e^x

 

[icystrike]

yes. i understand . However, why did they take M(e)=1?
I can't really accept their explanation : M(2)<1 and M(4)>1 ,thus they allow M(e)=1..
My question is how about other real number between 2 to 4?

 

[HallsofIvy]

It's NOT so much "how" as "why". I presume that just before the section you quoted they showed that the derivative of e^x, for e any number, "M(e)e^x" where M(e) is a number depending on e only, not on x.

It is not too difficult to show that M(2) is less than one and that M(3) is larger than 1 (by numerically approximating the limits). "Choosing" M(e) to be 1 is really choosing a specific value of e such that M(e)= 1. My point about M(2) and M(3) is that there is such a value of e, between 2 and 3. You could, then, by a succesion of numerical approximations, show that M(2.7) is less than 1 but that M(2.8) is greater than 1 so "e" is between 2.7 and 2.8. Or that M(2.71) is less than 1 but that M(2.72) is greater than one so that "e" is between 2.71 and 2.72, etc.

Choosing "e" to be the number such that M(e)= 1 means that the derivative of the function y= e^x is just e^x again.

 

[icystrike]

Oh! Thanks hallsofivy.(= I was rather taken aback by the fact that M(e)=1 and not any other real numbers between probably 2.71 to 2.72 . once again thanks!

 

[rzaidan]

Hi icystrike ;
M(e)=1 since e is defined to be the real number such that the area under the function
f(x)=1/x and x=1 ,x=e and the x-axis is equal to 1.
Best Wishes
Riad Zaidan