Is e^x Always Greater Than x for All Real Numbers?

[madah12]

Homework Statement

proving it

Homework Equations

The Attempt at a Solution

from (-infinity , 0) x is <0
e^x >0
for if there exist an positive a such that e^x=-a
then x =ln(-a) which is undefined
therefore e^x>0>x in (-infinity , 0)
from [0,infinity) since x+1>x we prove that e^x>x+1>x
e^(0)=1
0+1=1
d/dx e^x=e^x
d/dx (x+1)=1
e^x>1 for every x >0
f(x)=e^x , g(x)=x+1
f(a)=g(a) , f'(a)>g'(a) so f(b)>g(b) for every b in (0,infinity)
and since e^x>x+1 in (0,infinity) and x+1>x in R then e^x >x+1>x in (0,infinity)
therefore e^x >x for all x in R
-----------------------------------------------------------------------------
I am thinking it sounds stupid and there must be a better way to prove it?

 

[LeonhardEuler]

The Taylor series could come in handy for x>0.

 

[jgens]

The Taylor series could come in handy for x>0.

Unless you've defined the exponential function in terms of a Taylor series, I think that the OPs approach is simpler. Just show that exp(0) > 0 and then show that exp'(x) > 1 if x > 0.

 

[LeonhardEuler]

It's common to define the exponential that way. Then you just have
e^x = 1 + x + (positive terms)
and you're done.

 

[jgens]

True. And this method works well if that's the particular definition of the exponential function that you're working with. Most elementary calculus book that I've seen don't define the exponential function in this way however, and that's why I think that an alternative way might be more suitable for the OP.

 

[hunt_mat]

What about proving the function exp(x)-x in and increasing function? Let f(x)=exp(x)-x. Then f'(x)= e^{x}-1 which is positive for x bigger or equal to zero. So f is an inreasing function for x >=0. Evaluate at x=0 to show:
<br /> e^{x}-x\geqslant e^{0}-0=1&gt;0\Rightarrow e^{x}&gt;x<br />

 

[madah12]

What about proving the function exp(x)-x in and increasing function? Let f(x)=exp(x)-x. Then f'(x)= e^{x}-1 which is positive for x bigger or equal to zero. So f is an inreasing function for x >=0. Evaluate at x=0 to show:
<br /> e^{x}-x\geqslant e^{0}-0=1&gt;0\Rightarrow e^{x}&gt;x<br />

well yes but that only covers the positive part and not the negative on.

 

[jgens]

Suppose exp(a) < 0 for some a < 0. Choose b such that a+b=0. Clearly b > 0, so exp(b) > 0. Now consider exp(0) = exp(a+b) = exp(a)exp(b) and arrive at a contradiction.

 

[madah12]

Suppose exp(a) < 0 for some a < 0. Choose b such that a+b=0. Clearly b > 0, so exp(b) > 0. Now consider exp(0) = exp(a+b) = exp(a)exp(b) and arrive at a contradiction.

Oh thank s very much ,but it unusual for you to people to give immediate step by step solutions so how come?

 

[whyyie]

Well, I am not sure whether this can be consider as proving.
Can someone assure me?

First we assume e^x = x.
then we sketch 2 graphs:
y= e^x
y=x

From the graph, e^x > x for all x in R. Therefore, e^x > x for all x in R.

 

[Dickfore]

You may use the fact that the second derivative:

<br /> (e^{x})&#039;&#039; = e^{x} &gt; 0<br />

is positive, which means that the function is convex downwards. This means it lies above any of its tangents. Try to find the tangent that passes through the origin. You will then have a more rigorous inequality than the one you are trying to prove.

 

[Mark44]

Well, I am not sure whether this can be consider as proving.
Can someone assure me?

First we assume e^x = x.
then we sketch 2 graphs:
y= e^x
y=x

From the graph, e^x > x for all x in R. Therefore, e^x > x for all x in R.

Just showing the graphs of the two functions would not be considered a proof, IMO. Several methods have already been discussed in this thread, including the one that hunt_mat suggest about showing that f(x) = e^x - x is an increasing function.

 

[JonF]

Wouldn’t it just be easier to prove e^x-x is positive everywhere by looking at the global min. Taking the dervative of this function should be easy.

 

[hunt_mat]

well yes but that only covers the positive part and not the negative on.

so let x=-y with y>0 and repeat the process again.

 

[icystrike]

so let x=-y with y>0 and repeat the process again.

Hi there... the previous part of the proof will be valid to proof for x as all real number provided we assume:

s=e^{x} -x

when x--> -inf , s= inf (max)

ds/dx= e^{x}-1

ds/dx=0 when x=0 (min pt)