Proving the Identity for cosh Using Exponential Function Properties

[aaaa202]

Show that cosh(x) = 1 => x = 0

I am only allowed to use the definition of cosh, the algebraic rules for the exponential function, that exp(x2)>exp(x1) for x2 > x1, and the fact that we have defined it with the requirement:

exp(x) ≥ 1 + x

The exp(x) term of course is not trouble.

What I lack prooving is that:
exp(x) > 1 - x

Any ideas how I might approach it? I don't need answers, just hints

 

[CompuChip]

If x > 0, then 2x > 0 so 1 + x > 1 + x - 2x.

If x < 0, then it's not true (exp goes to 0 while 1 - x goes to infinity for x \to -\infty).

 

[aaaa202]

oh sorry I can see I made a fatal error.
I meant i need to show:
exp(-x) > 1 - x

 

[Mark44]

oh sorry I can see I made a fatal error.
I meant i need to show:
exp(-x) > 1 - x

A quick sketch of the graphs of y = e^-x and y = 1 - x shows that e^-x ≥ 1 - x for all real x, and that we get equality at only one value of x.

Assuming you can use techniques of calculus, let f(x) = e^-x - (1 - x), and find the minimum value and where it occurs.

 

[aaaa202]

I know calculus but this is an analysis course, where it has not yet been introduced. Thus I am to find another way of proving it.

 

[Mark44]

I know calculus but this is an analysis course, where it has not yet been introduced.

Really? Are you saying that you haven't learned differentiation, and you're in an analysis course?

Thus I am to find another way of proving it.

 

[aaaa202]

I'm talking a rigorous analysis course where everything is built up slowly. So far we have covered continuity and the properties of limits. I know differential calculus, but I am not allowed to use it for this handout.

 

[vela]

If you know that $e^x \ge 1+x$ for all $x$, you can plug anything you want in for $x$. In particular, you can plug in $-x$.